If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it. - Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

Word Problems on Combinatorics

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.
Use the functions in your TI-84 Plus or TI-Nspire to solve some of the questions in order to save time.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Pre-requisites:
(1.) Combinatorics: Example 1
(2.) Combinatorics: Example 2

Students may use:
(1.) Formulas
(2.) Calculators

Solve all questions.
Use at least two methods whenever applicable.
State the case/type of combinatorics for each question.
Show all work.

(1.) ACT In his costume supplies, Elmo the clown has 4 noses, 3 pairs of lips, and 2 wigs.
A clown costume consists of 1 nose, 1 pair of lips, and 1 wig.
How many different clown costumes can Elmo make?

First Approach: Fundamental Counting Principle

$ Number\;\;of\;\;different\;\;clown\;\;costumes \\[3ex] = 4 * 3 * 2 \\[3ex] = 24\;costumes \\[5ex] $ Second Approach: Combination

$ Number\:\:of\:\:different\;\;clown\;\;costumes \\[3ex] = C(4, 1) * C(3, 1) * C(2, 1) \\[3ex] C(n, r) = \dfrac{n!}{(n - r)!r!} \\[5ex] C(4, 1) \\[3ex] = \dfrac{4!}{(4 - 1)! * 1!} \\[5ex] = \dfrac{4 * 3!}{3! * 1} \\[5ex] = 4 \\[3ex] C(3, 1) \\[3ex] = \dfrac{3!}{(3 - 1)! * 1!} \\[5ex] = \dfrac{3 * 2!}{2! * 1} \\[5ex] = 3 \\[3ex] C(2, 1) \\[3ex] = \dfrac{2!}{(2 - 1)!1!} \\[5ex] = \dfrac{2 * 1!}{2! * 1} \\[5ex] = 2 \\[3ex] \implies \\[3ex] Number\;\;of\;\;different\;\;clown\;\;costumes \\[3ex] = 4 * 3 * 2 \\[3ex] = 24\;costumes \\[3ex] $

(2.) ACT In how many distinct orders can 5 students stand in line to buy yearbooks?

$ A.\:\: 5 \\[3ex] B.\:\: 15 \\[3ex] C.\:\: 25 \\[3ex] D.\:\: 120 \\[3ex] E.\:\: 3,125 \\[3ex] $

Permutation

$ Number\:\:of\:\:distinct\:\:orders \\[3ex] = 5! \\[3ex] = 5 * 4 * 3 * 2 * 1 \\[3ex] = 120\:\:orders $

(3.) ACT At a school picnic, 1 junior and 1 senior will be selected to lead the activities.
If there are 125 juniors and 100 seniors at the picnic, how many different 2-person combinations of 1 junior and 1 senior are possible?

$ F.\:\: 25 \\[3ex] G.\:\: 100 \\[3ex] H.\:\: 125 \\[3ex] J.\:\: 225 \\[3ex] K.\:\: 12,500 \\[3ex] $ Before you attempt this question, did you notice Question (1.)?

First Approach: Fundamental Counting Principle

$ Number\:\:of\:\:2-person\:\:combinations \\[3ex] = 125 * 100 \\[3ex] = 12500\:combinations \\[5ex] $ Second Approach: Combination

$ Number\:\:of\:\:2-person\:\:combinations = C(125, 1) * C(100, 1) \\[3ex] C(n, r) = \dfrac{n!}{(n - r)!r!} \\[5ex] C(125, 1) \\[3ex] = \dfrac{125!}{(125 - 1)! * 1!} \\[5ex] = \dfrac{125!}{124! * 1} \\[5ex] = \dfrac{125 * 124!}{124!} \\[5ex] = 125 \\[3ex] C(100, 1) \\[3ex] = \dfrac{100!}{(100 - 1)!1!} \\[5ex] = \dfrac{100!}{99! * 1} \\[5ex] = \dfrac{100 * 99!}{99!} \\[5ex] = 100 \\[3ex] Number\:\:of\:\:2-person\:\:combinations = 125 * 100 = 12500\:combinations $

(4.) ACT Happy Soup Company stamps a 6-character product code on each can of soup it produces.
Each product code consists of 5 letters (from the 26-letter alphabet) followed by a single digit (from the digits 0 to 9).
The letters may repeat.
How many such product codes are possible?

$ F.\:\: 5(26)(10) \\[3ex] G.\:\: 5(4)(3)(2) \\[3ex] H.\:\: 1^5(10) \\[3ex] J.\:\: 26(25)(24)(23)(22)(10) \\[3ex] K.\:\: 26^5(10) \\[3ex] $

Fundamental Counting Principle
$6-character$ product code.
$5$ letters and $1$ digit.
The letters may repeat.
$\underline{26}$ $\underline{26}$ $\underline{26}$ $\underline{26}$ $\underline{26}$ $\underline{10}$

$ Number\:\:of\:\:product\:\:codes \\[3ex] = 26 * 26 * 26 * 26 * 26 * 10 \\[3ex] = 26^5(10) $

(5.) A group of five friends went to an African restaurant for dinner.
The server greeted them, and seated them accordingly.
The restaurant offers seven different main dishes.

(a.) Suppose that the group collectively orders three different main dishes to share.
The server needs to place all three dishes in the center of the table.
How many different possible orders are there for the group?

(b.) Suppose that each person orders a main dish.
The server must remember who ordered which dish as part of the order.
It is possible for more than one person to order the same dish.
How many different possible orders are there for the group?

$ (a.) \\[3ex] \underline{Combination} \\[3ex] C(n, r) = \dfrac{n!}{(n - r)!r!} \\[5ex] Number\;\;of\;\;different\;\;orders \\[3ex] n = 7 \\[3ex] r = 3 \\[3ex] 7\;\;Combination\;\;3 \\[3ex] C(7, 3) \\[3ex] = \dfrac{7!}{(7 - 3)! * 3!} \\[5ex] = \dfrac{7!}{4! * 3!} \\[5ex] = \dfrac{7 * 6 * 5 * 4!}{4! * 3 * 2 * 1} \\[5ex] = 7 * 5 \\[3ex] = 35\;\;orders \\[5ex] (b.) \\[3ex] Any\;\;of\;\;the\;\;5\;\;friends\;\;can\;\;order\;\;any\;\;of\;\;the\;\;7\;\;main\;\;dishes \\[3ex] Number\;\;of\;\;different\;\;orders \\[3ex] \underline{Fundamental\;\;Counting\;\;Principle} \\[3ex] \underline{Friend\;1}\:\:\:\underline{Friend\;2}\:\:\:\underline{Friend\;3}\:\:\:\underline{Friend\;4}\:\:\:\underline{Friend\;5} \\[3ex] \;\;\; 7 \;\;\;\;\;\;\; * \;\;\;\;\;\;\; 7 \;\;\;\;\;\;\; * \;\;\;\;\; 7 \;\;\;\;\; * \;\;\;\;\; 7 \;\;\;\;\; * \;\;\;\;\;\; 7 \\[3ex] = 7^5 \\[3ex] = 16807\;\;orders $

(6.) JAMB In how many ways can 2 students be selected from a group of 5 students in a debating competition?

$ A.\:\: 25\:\:ways \\[3ex] B.\:\: 20\:\:ways \\[3ex] C.\:\: 15\:\:ways \\[3ex] D.\:\: 10\:\:ways \\[3ex] $

No order...any $2$ students from any $5$ students

Combination

$ C(n, r) = \dfrac{n!}{(n - r)!r!} \\[5ex] Number\:\:of\:\:ways \\[3ex] = C(5, 2) \\[3ex] = \dfrac{5!}{(5 - 2)! * 2!} \\[5ex] = \dfrac{5!}{3! * 2!} \\[5ex] = \dfrac{5 * 4 * 3!}{3! * 2 * 1} \\[5ex] = 5 * 2 \\[3ex] = 10\:\:ways $

(7.) Deborah needs to print her thesis.
Her thesis is a hundred and fifty pages.
Each page must be printed only one time.
With her school ID, she can use any of the ten printers in the library to print her thesis.

(a.) How many ways can she print her thesis using any of the printers?

Assume only three of the printers can print both colored and black/white pages.
Pages $30 - 34$ of Deborah's thesis contain high-definition pictures.
Deborah needs to print these specific pages in color.

(b.) How many ways can she print her thesis using all the printers?

Assume only three of the printers can print only colored pages.
Pages $30 - 34$ of Deborah's thesis contain high-definition pictures.
Deborah needs to print these specific pages in color.

(c.) How many ways can she print her thesis using all the printers?

(d.) Assume all the printers can print in black/white.
Deborah later decides to print the entire thesis in black and white.
However, the librarian informed her that each printer can only print a set of $30$ consecutive pages at a time for the same student.
How many ways can she print her thesis using all the printers?

Fundamental Counting Principle

(a.)
Any of the ten printers can print the first page.
So, there are ten ways to print the first page.

Similarly, any of the ten printers can print the second page.

So, there are ten ways to print the second page.

So, each of the $150$ pages can be printed by any of the ten printers.
This means that each of the $150$ pages can be printed in ten ways.

$ Number\;\;of\;ways \\[3ex] Page\;1\;\;\;Page\;2\;\;\;Page\;3\;\;\;...\;\;\;Page\;148\;\;\;Page\;149\;\;\;Page\;150 \\[3ex] \underline{10}\;\;\;\:\:\:\;\;\;\;\;\;\underline{10}\;\;\;\:\:\:\;\;\;\;\;\;\underline{10}\;\;\;\;\;\;...\;\;\;\;\;\;\underline{10}\;\;\;\:\:\:\;\;\;\;\;\;\underline{10}\;\;\;\:\:\:\;\;\;\;\;\;\underline{10} \\[3ex] = 10 * 10 * 10 * ... * 10 * 10 * 10 \\[3ex] = 10^{150}\:\:ways \\[3ex] $ (b.)
Three printers can print in color and black/white.
Pages $30 - 34$ needs to be printed in color.
This means that $5$ pages ($30, 31, 32, 33, 34$) must be printed in color.
Each of the $5$ pages can use any of those three printers.
The remaining $150 - 5 = 145\;\;pages$ can be printed using any of the ten printers.
This is because those three printers can print in both color and black/white.
So:
Each of the $145$ pages can be printed using any of the ten printers.

$ Number\;\;of\;\;ways \\[3ex] Page\;1\;\;\;Page\;2\;\;\;Page\;3\;\;\;...\;\;\;Page\;30\;\;\;Page\;31\;\;\;Page\;32\;\;\;Page\;33\;\;\;Page\;34\;\;\;...\;\;\;Page\;148\;\;\;Page\;149\;\;\;Page\;150 \\[3ex] \underline{10}\;\;\;\:\:\:\;\;\;\;\;\;\underline{10}\;\;\;\:\:\:\;\;\;\;\;\;\underline{10}\;\;\;\;\;\;...\;\;\;\;\;\;\underline{3}\;\;\;\;\;\;\:\:\:\;\;\;\;\;\;\underline{3}\;\;\;\;\;\;\;\:\:\:\;\;\;\;\;\;\underline{3}\;\;\;\;\;\;\;\:\:\:\;\;\;\;\underline{3}\;\;\;\;\;\;\:\:\:\;\;\;\;\;\;\underline{3}\;\;\;\;\;\;\;\;...\;\;\;\;\;\;\underline{10}\;\;\;\:\:\:\;\;\;\;\;\;\underline{10}\;\;\;\:\:\:\;\;\;\;\;\;\underline{10} \\[3ex] = 10 * 10 * 10 * ... * 10 * 10 * 3 * 3 * 3 * 3 * 3 * 10 * 10 * ... * 10 * 10 * 10 \\[3ex] = 243 * 10^{145}\;\;ways \\[3ex] $ (c.)
Three printers can print in color and black/white.
Pages $30 - 34$ needs to be printed in color.
This means that $5$ pages ($30, 31, 32, 33, 34$) must be printed in color.
Each of the $5$ pages can use any of those three printers.
The remaining $150 - 5 = 145\;\;pages$ can be printed using any of the remaining ($10 - 3 = 7$) printers.
This is because those three printers can only print in color.
So:
Each of the $145$ pages can be printed using any of the seven printers.

$ Number\;\;of\;\;ways \\[3ex] Page\;1\;\;\;Page\;2\;\;\;Page\;3\;\;\;...\;\;\;Page\;30\;\;\;Page\;31\;\;\;Page\;32\;\;\;Page\;33\;\;\;Page\;34\;\;\;...\;\;\;Page\;148\;\;\;Page\;149\;\;\;Page\;150 \\[3ex] \;\;\;\;\underline{7}\;\;\;\;\;\;\;\:\:\:\;\;\underline{7}\;\;\;\;\;\;\;\:\:\:\;\;\;\underline{7}\;\;\;\;\;\;\;\;...\;\;\;\;\;\;\;\underline{3}\;\;\;\;\;\;\;\;\;\:\:\:\;\underline{3}\;\;\;\;\;\;\;\;\;\;\:\:\:\;\;\underline{3}\;\;\;\;\;\;\;\;\;\;\:\:\:\;\underline{3}\;\;\;\;\;\;\;\;\;\:\:\:\;\;\;\underline{3}\;\;\;\;\;\;\;\;\;\;\;...\;\;\;\;\;\;\;\;\underline{7}\;\;\;\;\;\;\:\:\:\;\;\;\;\;\;\;\underline{7}\;\;\;\:\:\:\;\;\;\;\;\;\;\;\underline{7} \\[3ex] = 7 * 7 * 7 * ... * 7 * 7 * 3 * 3 * 3 * 3 * 3 * 7 * 7 * ... * 7 * 7 * 7 \\[3ex] = 243 * 7^{145}\;\;ways \\[3ex] $ (d.)
Each printer can only print a set of $30$ consecutive pages at a time

$ 30\;pages = 1\;set \\[3ex] 150\;pages = 150 \div 30 = 5\;sets \\[3ex] $ This means that any of the ten printers can print any $5$ sets.

$ Number\;\;of\;ways \\[3ex] \underline{Pages\;1 - 30}\:\:\:\underline{Pages\;31 - 60}\:\:\:\underline{Pages\;61 - 90}\:\:\:\underline{Pages\;91 - 120}\:\:\:\underline{Pages\;121 - 150} \\[3ex] 10 \;\;\;\;\;\;\;\;\;\;\;\; * \;\;\;\;\;\;\;\;\;\;\;\; 10 \;\;\;\;\;\;\;\;\;\;\;\; * \;\;\;\;\;\;\;\;\;\;\;\; 10 \;\;\;\;\;\;\;\;\;\;\;\; * \;\;\;\;\;\;\;\;\;\;\;\; 10 \;\;\;\;\;\;\;\;\;\;\;\; * \;\;\;\;\;\;\;\;\;\;\;\; 10 \\[3ex] = 10^5\;\;ways $

(8.) (a.) How many different signals can be made by hoisting three yellow flags, four green flags, and two red flags on a ship's mast at the same time?

(b.) The custom license plate of a state has five characters.
The characters can be a combination of uppercase letters and non-zero digits.
Determine the number of custom license plates that begin with two uppercase letters and end with three digits if the letter or digit cannot be repeated?

(c.) Fifteen applicants applied for four open positions in a company.
How many ways can the positions be filled if no preference is given to any applicant?

(d.) Fifteen applicants applied for four open positions in a company.
How many ways can the positions be filled if a particular applicant must be given a position?

(e.) A binary string has a length of four.
Write the binary string.
Determine the number of binary strings that has only alternate bits ($0101$ or $1010$) somewhere in the string?
Determine the number of binary strings that has at least two adjacent bits that are the same ($00$ or $11$) somewhere in the string?

$ (a.) \\[3ex] \underline{Permutation\;\;with\;\;Duplicates} \\[3ex] Let: \\[3ex] Yellow = Y \\[3ex] Green = G \\[3ex] Red = R \\[3ex] S = \{Y, Y, Y, G, G, G, G, R, R\} \\[3ex] S = \{3\;Yellow, 4\;Green, 2\;Red\} \\[3ex] n(S) = 3 + 4 + 2 = 9 \\[3ex] Number\;\;of\;\;signals \\[3ex] = \dfrac{9!}{3! * 4! * 2!} \\[5ex] = \dfrac{9 * 8 * 7 * 6 * 5 * 4!}{3 * 2 * 1 * 4! * 2 * 1} \\[5ex] = 9 * 4 * 7 * 5 \\[3ex] = 1260\;signals \\[5ex] (b.) \\[3ex] Two\;\;ways\;\;to\;\;solve\;\;it \\[3ex] Let:\;\; Uppercase\;\;Letter = L \\[3ex] L = \{A, B, C, D, ..., X, Y, Z\} \\[3ex] n(L) = 26 \\[3ex] Mon-zero\;\;Digit = D \\[3ex] D = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] n(D) = 9 \\[3ex] No\;\;repetition \\[3ex] \underline{1st\;\;Method:\;\;Multiplication\;\;Principle} \\[3ex] Number\;\;of\;\;license\;\;plates \\[3ex] \underline{1st}\:\:\:\underline{2nd}\:\:\:\underline{3rd}\:\:\:\underline{4th}\:\:\:\underline{5th} \\[3ex] 26 \;\; * \;\; 25 \;\; * \;\; 9 \;\; * \;\; 8 \;\; * \;\; 7 \\[3ex] = 327600\;\;license\;\;plates \\[3ex] \underline{2nd\;\;Method:\;\;Permutation} \\[3ex] P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] Number\;\;of\;\;license\;\;plates \\[3ex] = Perm\;\;2\;\;from\;\;26 * Perm\;\;3\;\;from\;\;9 \\[5ex] = P(26, 2) * P(9, 3) \\[3ex] = \dfrac{26!}{(26 - 2)!} * \dfrac{9!}{(9 - 3)!} \\[5ex] = \dfrac{26!}{24!} * \dfrac{9!}{6!} \\[5ex] = \dfrac{26 * 25 * 24!}{24!} * \dfrac{9 * 8 * 7 * 6!}{6!} \\[5ex] = 26 * 25 * 9 * 8 * 7 \\[3ex] = 327600\;\;license\;\;plates \\[5ex] (c.) \\[3ex] \underline{Combination} \\[3ex] C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] n = 15 \\[3ex] r = 4 \\[3ex] Number\;\;of\;\;positions \\[3ex] = C(15, 4) \\[3ex] = \dfrac{15!}{(15 - 4)! * 4!} \\[5ex] = \dfrac{15!}{11! * 4!} \\[5ex] = \dfrac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1} \\[5ex] = 15 * 7 * 13 \\[3ex] = 1365\;\;positions \\[5ex] (d.) \\[3ex] \underline{Combination} \\[3ex] 1\;\;applicant\;\;must\;\;be\;\;given\;\;a\;\;position \\[3ex] Select\;\;1\;\;from\;\;1 \\[3ex] Remaining\;\;Applicants:\;\;n = 15 - 1 = 14 \\[3ex] Remaining\;\;positions:\;\; r = 4 - 1 = 3 \\[3ex] Select\;\;3\;\;from\;\;14 \\[5ex] Number\;\;of\;\;positions \\[3ex] = C(1, 1) * C(14, 3) \\[3ex] = \dfrac{1!}{(1 - 1)! * 1!} * \dfrac{14!}{(14 - 3)! * 3!} \\[5ex] = \dfrac{1!}{0! * 1!} * \dfrac{14!}{11! * 3!} \\[5ex] = \dfrac{1}{1 * 1} * \dfrac{14 * 13 * 12 * 11!}{11! * 3 * 2 * 1} \\[5ex] = 1 * 14 * 13 * 2 \\[3ex] = 1365\;\;positions \\[5ex] (e.) \\[3ex] Binary\;\;consists\;\;of\;\;only\;\;0's\;\;and\;\;1's \\[3ex] Length\;\;of\;\;4 \\[3ex] Total\;\;number\;\;of\;\;binary\;\;strings = 2^4 = 16\;\;strings \\[3ex] They\;\;are\;\; \\[3ex] 0000 \\[3ex] 0001 \\[3ex] 0010 \\[3ex] 0011 \\[3ex] 0100 \\[3ex] 0101 \\[3ex] 0110 \\[3ex] 0111 \\[3ex] 1000 \\[3ex] 1001 \\[3ex] 1010 \\[3ex] 1011 \\[3ex] 1100 \\[3ex] 1101 \\[3ex] 1110 \\[3ex] 1111 \\[3ex] 2\;\;strings\;\;have\;\;only\;\;alternate\;\;bits \\[3ex] They\;\;are:\;\; 0101\;\;and\;\;1010 \\[3ex] The\;\;rest\;\;has\;\;at\;\;least\;\;two\;\;adjacent\;\;same\;\;bits \\[3ex] 16 - 2 = 14 \\[3ex] \therefore\;\; 14\;\;strings\;\;have\;\;at\;\;least\;\;two\;\;adjacent\;\;same\;\;bits $

(9.) NSC Given the digits: 3; 4; 5; 6; 7; 8 and 9
Calculate how many unique 5-digit codes can be formed using the digits above, if:
(5.1) The digits may be repeated
(5.2) The digits may not be repeated

$ 3;4;5;6;7;8;9 \\[3ex] total = 7 \\[3ex] unique\:\:5-digit\:\:codes \\[3ex] some = 5 \\[3ex] (5.1) \\[3ex] \underline{Fundamental\:\:Counting\:\:Principle} \\[3ex] Number\:\:of\:\:unique\:\:5-digit\:\:codes \\[3ex] \underline{3 - 9}\:\:\:\underline{3 - 9}\:\:\:\underline{3 - 9}\:\:\:\underline{3 - 9}\:\:\:\underline{3 - 9} \\[3ex] \;\;\; 7 \;\; * \;\; 7 \;\; * \;\; 7 \;\; * \;\; 7 \;\; * \;\; 7 \\[3ex] = 16807\:\:codes \\[3ex] OR \\[3ex] Number\:\:of\:\:unique\:\:5-digit\:\:codes \\[3ex] = total^{some} \\[3ex] = 7^5 \\[3ex] = 16807\:\:codes \\[3ex] (5.2) \\[3ex] \underline{Permutation} \\[3ex] P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] Number\:\:of\:\:unique\:\:5-digit\:\:codes \\[3ex] = P(total,some) \\[3ex] = P(7, 5) \\[3ex] = \dfrac{7!}{(7 - 5)!} \\[5ex] = \dfrac{7!}{2!} \\[5ex] = \dfrac{7 * 6 * 5 * 4 * 3 * 2!}{2!} \\[5ex] = 7 * 6 * 5 * 4 * 3 \\[3ex] = 2520\:\:codes $

(10.) Thirty people purchase raffle tickets.
Three winning tickets are selected at random.
In how many different ways can the prizes be awarded if the first prize is $5000, the second prize is $3000, and the third prize is $1000?

This deals with order...first prize, second prize, third prize

Permutation

$ P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] Number\:\:of\:\:ways \\[3ex] = P(30, 3) \\[3ex] = \dfrac{30!}{(30 - 3)!} \\[5ex] = \dfrac{30!}{27!} \\[5ex] = \dfrac{30 * 29 * 28 * 27!}{27!} \\[5ex] = 30 * 29 * 28 \\[3ex] = 24,360\:\:ways $

(11.) NSC Given the digits: 3; 4; 5; 6; 7; 8 and 9
How many unique 3-digit codes can be formed using the digits above, if:

Digits may be repeated
The code is greater than 400 but less than 600
The code is divisible by 5

$ 3;4;5;6;7;8;9 \\[3ex] total = 7 \\[3ex] unique\:\:3-digit\:\:codes\:\:with\:\:several\:\:conditions \\[3ex] some = 3 \\[3ex] \underline{Fundamental\:\:Counting\:\:Principle} \\[3ex] $ 1st condition: Digits may be repeated.

$ \underline{3 - 9} \hspace{5ex} \underline{3 - 9} \hspace{4ex} \underline{3 - 9} \\[3ex] \hspace{2ex} 7 \hspace{3ex} * \hspace{4ex} 7 \hspace{3ex} * \hspace{3ex} 7 \\[3ex] $ 2nd condition: The code is greater than 400 but less than 600
This means that the first digit of the code must be 4 or 5
Any number of 4 or 5 must be the first digit.
This means that the first digit can be written in any two ways

$ \underline{4 - 5} \hspace{5ex} \underline{3 - 9} \hspace{4ex} \underline{3 - 9} \\[3ex] \hspace{2ex} 2 \hspace{3ex} * \hspace{4ex} 7 \hspace{3ex} * \hspace{3ex} 7 \\[3ex] $ 3rd condition: The code is divisible by 5
Any number divisible by 5 must end in 0 or 5
There is no 0 in the digit.
There is only 5

This means that the last digit of the code must be 5
This means that the last digit can be written in only one way

$ \underline{4 - 5} \hspace{5ex} \underline{3 - 9} \hspace{5ex} \underline{5} \\[3ex] \hspace{2ex} 2 \hspace{3ex} * \hspace{4ex} 7 \hspace{3ex} * \hspace{2ex} 1 \\[3ex] $ Number of unique 3-digit codes = 14 codes

(12.) Thirty people purchase raffle tickets.
Three winning tickets are selected at random.
In how many different ways can the prizes be awarded if each prize is $1000?

This does not deal with order...any/each prize is $\$1000$

Combination

$ C(n, r) = \dfrac{n!}{(n - r)!r!} \\[5ex] Number\:\:of\:\:ways \\[3ex] = C(30, 3) \\[3ex] = \dfrac{30!}{(30 - 3)! * 3!} \\[5ex] = \dfrac{30!}{27! * 3!} \\[5ex] = \dfrac{30 * 29 * 28 * 27!}{27! * 3 * 2 * 1} \\[5ex] = 5 * 29 * 28 \\[3ex] = 4,060\:\:ways $

(13.) ACT A company prints contest codes on its fun-size bags of candy.
Each 6-character code consists of the letter A followed by the letter H followed by 4 of the digits 0 through 9
The digits may repeat.
Which of the following expressions gives the number of different 6-character codes that are possible?

$ F.\;\; 1(1)(10)(10)(10)(10) \\[3ex] G.\;\; 2(1)(10)(9)(8)(7) \\[3ex] H.\;\; 2(1)(10)(10)(10)(10) \\[3ex] J.\;\; 2(2)(10)(9)(8)(7) \\[3ex] K.\;\; 2(2)(10)(10)(10)(10) \\[3ex] $

Fundamental Counting Principle

Each $6-character$ code consists of the letter $A$
This can be done in only one way.
followed by the letter $H$
This can also be done in only one way.
followed by $4$ of the digits $0$ through $9$
The digits may repeat.
$0 - 9$ is ten digits
So, each of the $4$ digits (remaining four characters of the 6-character code) can be any of the $10$ digits because the digits may repeat.

$ 6-character\;\;code \\[3ex] \underline{A}\;\;\;\;\;\;\underline{H}\;\;\;\;\;\;\underline{0 - 9}\;\;\;\underline{0 - 9}\;\;\;\underline{0 - 9}\;\;\;\underline{0 - 9} \\[3ex] 1 \;\;*\;\; 1 \;\;\;*\;\;\; 10 \;\;\;*\;\;\; 10 \;\;\;*\;\;\; 10 \;\;\;*\;\;\; 10 \\[3ex] = 1(1)(10)(10)(10)(10) \;\;different\;\;6-character\;\;codes $

(14.) ACT In a window display at a flower shop, there are 3 spots for 1 plant each.
To fill these 3 spots, Adam has 7 plants to select from, each of a different type.
Selecting from the 7 plants, Adam can make how many possible display arrangements with 1 plant in each spot?
(Note: The positions of the unselected plants do not matter.)

$ F.\;\; 3 \\[3ex] G.\;\; 7 \\[3ex] H.\;\; 18 \\[3ex] J.\;\; 210 \\[3ex] K.\;\; 343 \\[3ex] $

Fundamental Counting Principle and Permutation

Fundamental Counting Principle

$3$ spots for $1$ plant
Any $1$ of $7$ different plants can be selected to fill the first spot.
This means that the first spot can be filled by any one of the seven plants.
This means that the first spot can be filled in seven ways.

Once the first spot is filled by one plant, two other spots remain for the remaining six plants.
The second spot can be filled by any one of the six remaining plants.
This means that the second spot can be filled in six ways.

Once the second spot is filled, the third spot can be filled by any one of the remaining five plants.
This means that the third spot can be filled in five ways.

$ \underline{1st\:spot}\:\:\:\underline{2nd\:spot}\:\:\:\underline{3rd\:spot} \\[3ex] \;\;\;7 \:\:\:\:\:\:\;\;\; * \;\;\;\:\:\: 6 \;\;\;\:\:\: * \;\;\;\:\: 5 \\[3ex] = 210\;possible\;\;arrangements \\[3ex] $ The positions of the unselected plants do not matter.
However, the order does matter.
The order is the first, second, and third spots.
Any plant selected from the sevem plants can fill these spots.
However, the spots are ordered: first spot, second spot, third spot.
Therefore it is also Permutation

$ P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] Permutation\;\;of\;\;7\;plants\;\;taking\;\;3\;spots\;\;at\;\;a\;\;time \\[3ex] n = 7 \\[3ex] r = 3 \\[3ex] Number\;\;of\;\;arrangements \\[3ex] = P(7, 3) \\[3ex] = \dfrac{7!}{(7 - 3)!} \\[5ex] = \dfrac{7!}{4!} \\[5ex] = \dfrac{7 * 6 * 5 * 4!}{4!} \\[5ex] = 7 * 6 * 5 \\[3ex] = 210\;possible\;\;arrangements $

(15.) Esther helps her clients with creating, remembering, and storing passwords.
The password is a character that typically consists of digits, uppercase letters, lowercase letters and/or special characters.
There are ten digits: 0 - 9
There are twenty six uppercase letters of the English Language alphabet: A - Z
There are twenty six lowercase letters of the English Language alphabet: a - z
There are several special characters.
However, the special characters to be used for this question are six: !, @, #, %, &, *
The password length is seven.

(a.) How many passwords are possible if the password consists of digits, uppercase letters, lowercase letters and/or special characters that cannot be repeated?

(b.) How many passwords are possible if the password consists of digits, uppercase letters, lowercase letters and/or special characters that cannot be repeated and the first character cannot be a special character?

Some of her clients prefer a combination of only uppercase letters and digits
(c.) How many passwords are possible if the password must contain at least one digit?

(d.) How many passwords are possible if the password must contain at least one letter?

(e.) How many passwords are possible if the password must contain at least one digit and at least one letter?

Fundamental Counting Principle and
Permutation (if the characters do not repeat)

$ P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] 10\;\;digits = 10 \\[3ex] 26\;\;uppercase\;\;letters = 26 \\[3ex] 26\;\;lowercase\;\;letters = 26 \\[3ex] 6\;\;special\;\;characters = 6 \\[3ex] Sum = 10 + 26 + 26 + 6 = 68\;\;characters \\[3ex] $ Characters may not repeat.
For the first question:
Any of the $68$ characters can be used as the first character
Any of the remaining $67$ characters can be used as the second character
Any of the remaining $66$ characters can be used as the third character
Any of the remaining $65$ characters can be used as the fourth character
Any of the remaining $64$ characters can be used as the fifth character
Any of the remaining $63$ characters can be used as the sixth character
Any of the remaining $62$ characters can be used as the seventh character

$ (a.) \\[3ex] \underline{First\;\;Method:\;\;Fundamental\;\;Counting\;\;Principle} \\[3ex] Number\;\;of\;\;passwords \\[3ex] \underline{1st}\:\:\:\;\underline{2nd}\:\:\:\;\underline{3rd}\;\;\;\;\underline{4th}\;\;\;\underline{5th}\;\;\;\;\underline{6th}\;\;\;\;\underline{7th} \\[3ex] \; 68 \;*\; 67 \;*\; 66 \;*\; 65 \;*\; 64 \;*\; 63 \;*\; 62 \\[3ex] = 4885997276160\;\;passwords \\[3ex] \underline{Second\;\;Method:\;\;Permutation} \\[3ex] Perm\;\;7\;\;from\;\;68 \\[3ex] Number\;\;of\;\;passwords \\[3ex] = P(68, 7) \\[3ex] = \dfrac{68!}{(68 - 7)!} \\[5ex] = \dfrac{68!}{61!} \\[5ex] = \dfrac{68 * 67 * 66 * 65 * 64 * 63 * 62 * 61!}{61!} \\[5ex] = 68 * 67 * 66 * 65 * 64 * 63 * 62 \\[3ex] = 4885997276160\;\;passwords \\[3ex] $ For the second question:
The first character of the password cannot be a special character
Allowable characters for the first character of the password = $10 + 26 + 26 = 62$ characters
Characters may not repeat.
Once a character is used, it cannot be used again.
So, we can use any of the remaining $67$ characters for the remaning characters of the password.
Any of the $62$ characters can be used as the first character
Any of the remaining $67$ characters can be used as the second character
Any of the remaining $66$ characters can be used as the third character
Any of the remaining $65$ characters can be used as the fourth character
Any of the remaining $64$ characters can be used as the fifth character
Any of the remaining $63$ characters can be used as the sixth character
Any of the remaining $62$ characters can be used as the seventh character

$ (b.) \\[3ex] \underline{First\;\;Method:\;\;Fundamental\;\;Counting\;\;Principle} \\[3ex] Number\;\;of\;\;passwords \\[3ex] \underline{1st}\:\:\:\;\underline{2nd}\:\:\:\;\underline{3rd}\;\;\;\;\underline{4th}\;\;\;\underline{5th}\;\;\;\;\underline{6th}\;\;\;\;\underline{7th} \\[3ex] \; 62 \;*\; 67 \;*\; 66 \;*\; 65 \;*\; 64 \;*\; 63 \;*\; 62 \\[3ex] = 4454879869440\;\;passwords \\[3ex] \underline{Second\;\;Method:\;\;Permutation} \\[3ex] Perm\;\;1\;\;from\;\;62 * Perm\;\;6\;\;from\;\;67 \\[3ex] Number\;\;of\;\;passwords \\[3ex] = P(62, 1) * P(67, 6) \\[3ex] = \dfrac{62!}{(62 - 1)!} * \dfrac{67!}{(67 - 6)!} \\[5ex] = \dfrac{62!}{(61)!} * \dfrac{67!}{(61)!} \\[5ex] = \dfrac{62 * 61!}{(61)!} * \dfrac{67 * 66 * 65 * 64 * 63 * 62 * 61!}{(61)!} \\[5ex] = 62 * 67 * 66 * 65 * 64 * 63 * 62 \\[3ex] = 4454879869440\;\;passwords \\[3ex] $ For the third question:
Characters may repeat
Allowable characters = uppercase letters and digits = $26 + 10 = 36$
At least one digit means one or more digits
The best way to solve this kind of Combinatorics question (question that specifies "at least") is by using Complementary Rule of Probability
Probability of at least one is $1 - $ Porbability of none
Similarly, the number of possible passwords that must contain at least one digit is the total number of passwords that contains both uppercase letters and digits $-$ the number of passwords that contains no digits
No digits implies all uppercase letters
Therefore, the number of passwords that must contain at least one digit is the total number of passwords that contains both uppercase letters and digits $-$ the number of passwords that contains all uppercase letters

$ (c.) \\[3ex] \underline{Fundamental\;\;Counting\;\;Principle} \\[5ex] Total\;\;number\;\;of\;\;passwords\;\;containing\;\;uppercase\;\;letters\;\;and\;\;digits \\[3ex] \underline{1st}\:\:\:\;\underline{2nd}\:\:\:\;\underline{3rd}\;\;\;\;\underline{4th}\;\;\;\underline{5th}\;\;\;\;\underline{6th}\;\;\;\;\underline{7th} \\[3ex] \; 36 \;*\; 36 \;*\; 36 \;*\; 36 \;*\; 36 \;*\; 36 \;*\; 36 \\[3ex] = 36 * 36 * 36 * 36 * 36 * 36 * 36 \\[3ex] = 78364164096\;\;passwords \\[3ex] Number\;\;passwords\;\;containing\;\;only\;\;uppercase\;\;letters \\[3ex] \; 26 \;*\; 26 \;*\; 26 \;*\; 26 \;*\; 26 \;*\; 26 \;*\; 26 \\[3ex] = 26 * 26 * 26 * 26 * 26 * 26 * 26 \\[3ex] = 8031810176\;\;passwords \\[3ex] Number\;\;of\;\;passwords\;\;containing\;\;at\;\;least\;\;one\;\;digit \\[3ex] = 78364164096 - 8031810176 \\[3ex] = 70332353920\;\;passwords \\[3ex] $ For the fourth question:
Characters may repeat
Allowable characters = uppercase letters and digits = $26 + 10 = 36$
At least one letter means one or more letters
The best way to solve this kind of Combinatorics question (question that specifies "at least") is by using Complementary Rule of Probability
Probability of at least one is $1 - $ Porbability of none
Similarly, the number of possible passwords that must contain at least one uppercase letter is the total number of passwords that contains both uppercase letters and digits $-$ the number of passwords that contains no letters
No letters implies all digits
Therefore, the number of passwords that must contain at least one uppercase letter is the total number of passwords that contains both uppercase letters and digits $-$ the number of passwords that contains all digits

$ (d.) \\[3ex] \underline{Fundamental\;\;Counting\;\;Principle} \\[5ex] Total\;\;number\;\;of\;\;passwords\;\;containing\;\;uppercase\;\;letters\;\;and\;\;digits \\[3ex] \underline{1st}\:\:\:\;\underline{2nd}\:\:\:\;\underline{3rd}\;\;\;\;\underline{4th}\;\;\;\underline{5th}\;\;\;\;\underline{6th}\;\;\;\;\underline{7th} \\[3ex] \; 36 \;*\; 36 \;*\; 36 \;*\; 36 \;*\; 36 \;*\; 36 \;*\; 36 \\[3ex] = 36 * 36 * 36 * 36 * 36 * 36 * 36 \\[3ex] = 78364164096\;\;passwords \\[3ex] Number\;\;passwords\;\;containing\;\;only\;\;digits \\[3ex] \; 10 \;*\; 10 \;*\; 10 \;*\; 10 \;*\; 10 \;*\; 10 \;*\; 10 \\[3ex] = 10 * 10 * 10 * 10 * 10 * 10 * 10 \\[3ex] = 10000000\;\;passwords \\[3ex] Number\;\;of\;\;passwords\;\;containing\;\;at\;\;least\;\;one\;\;uppercase\;\;letter \\[3ex] = 78364164096 - 10000000 \\[3ex] = 78354164096\;\;passwords \\[3ex] $ For the fifth question:
Characters may repeat
Allowable characters = uppercase letters and digits = $26 + 10 = 36$
At least one digit means one or more digits
At least one letter means one or more letters
Reviewing our steps in the third and fourth questions:
To determine the number of passwords that must contain at least one digit and at least one letter, we need to subtract the sum of the number of passwords that contains all letters and the number of passwords that contains all digits from the total number of passwords containing both letters and digits

$ (e.) \\[3ex] \underline{Fundamental\;\;Counting\;\;Principle} \\[5ex] Total\;\;number\;\;of\;\;passwords\;\;containing\;\;uppercase\;\;letters\;\;and\;\;digits \\[3ex] \underline{1st}\:\:\:\;\underline{2nd}\:\:\:\;\underline{3rd}\;\;\;\;\underline{4th}\;\;\;\underline{5th}\;\;\;\;\underline{6th}\;\;\;\;\underline{7th} \\[3ex] \; 36 \;*\; 36 \;*\; 36 \;*\; 36 \;*\; 36 \;*\; 36 \;*\; 36 \\[3ex] = 36 * 36 * 36 * 36 * 36 * 36 * 36 \\[3ex] = 78364164096\;\;passwords \\[3ex] Number\;\;passwords\;\;containing\;\;only\;\;uppercase\;\;letters \\[3ex] \; 26 \;*\; 26 \;*\; 26 \;*\; 26 \;*\; 26 \;*\; 26 \;*\; 26 \\[3ex] = 26 * 26 * 26 * 26 * 26 * 26 * 26 \\[3ex] = 8031810176\;\;passwords \\[3ex] Number\;\;passwords\;\;containing\;\;only\;\;digits \\[3ex] \; 10 \;*\; 10 \;*\; 10 \;*\; 10 \;*\; 10 \;*\; 10 \;*\; 10 \\[3ex] = 10 * 10 * 10 * 10 * 10 * 10 * 10 \\[3ex] = 10000000\;\;passwords \\[3ex] Number\;\;of\;\;passwords\;\;containing\;\;at\;\;least\;\;one\;\;uppercase\;\;letter\;\;and\;\;at\;\;least\;\;one\;\;digit \\[3ex] = 78364164096 - (8031810176 + 10000000) \\[3ex] = 78364164096 - 8041810176 \\[3ex] = 70322353920\;\;passwords $

(16.) ACT Terri is a dress designer and owns a retail store.
She designs, produces, and sells 5 dress styles.
These styles and the production cost of 1 dress of each style are shown in the table below.

Style	Production cost of 1 dress
$A$	$\$15.00$
$B$	$\$25.00$
$C$	$\$45.00$
$D$	$\$60.00$
$E$	$\$65.00$

Terri will hang 1 dress of each style along a rod in the window of her store.
How many total possible orders (permutations) of these dresses are there for Terri to consider?

$ F.\;\; 1 \\[3ex] G.\;\; 5 \\[3ex] H.\;\; 25 \\[3ex] J.\;\; 120 \\[3ex] K.\;\; 3,125 \\[3ex] $

Fundamental Counting Principle

$ Number\;\;of\;\;ways \\[3ex] = 5! \\[3ex] = 5 * 4 * 3 * 2 * 1 \\[3ex] = 120\;\;ways \\[3ex] $ OR

Permutation

$ P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] Number\:\:of\:\:ways \\[3ex] = P(5, 5) \\[3ex] = \dfrac{5!}{(5 - 5)!} \\[5ex] = \dfrac{5!}{0!} \\[5ex] = \dfrac{5 * 4 * 3 * 2 * 1}{1} \\[5ex] = 120\:\:ways $

(17.) (a.) In how many different ways can a committee of twelve people be chosen from a group of twenty one politicians?

(b.) In how many different ways can a committee of twelve people be chosen from a group of sixteen Republicans and five Democrats?

(c.) In how many ways can a committee of twelve Republicans and two Democrats be chosen from a group of sixteen Republicans and five Democrats?

(d.) In how many different ways can a committee of twelve Republicans and two Democrats be chosen from a group of twenty one politicians?

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] (a.) \\[3ex] n = 21 \\[3ex] r = 12 \\[3ex] Number\;\;of\;\;ways \\[3ex] = C(21, 12) \\[3ex] = \dfrac{21!}{(21 - 12)! * 12!} \\[5ex] = \dfrac{21!}{9! * 12!} \\[5ex] = \dfrac{21 * 20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12!}{9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 * 12!} \\[5ex] = \dfrac{19 * 17 * 16 * 15 * 14 * 13}{8 * 6} \\[5ex] = \dfrac{19 * 17 * 2 * 15 * 14 * 13}{6} \\[5ex] = \dfrac{19 * 17 * 15 * 14 * 13}{3} \\[5ex] = 19 * 17 * 5 * 14 * 13 \\[3ex] = 293930\;\;ways \\[5ex] (b.) \\[3ex] Number\;\;of\;\;people \\[3ex] = 16\:\:Republicans + 5\;\;Democrats \\[3ex] = 21\;\;people \\[3ex] n = 21 \\[3ex] r = 12 \\[3ex] Same\;\;as\;\;(a.) \\[3ex] Number\;\;of\;\;ways \\[3ex] = C(21, 12) \\[3ex] = 293930\;\;ways \\[5ex] (c.) \\[3ex] 12\;\;Republicans\;\;from\;\;16\;\;Republicans \\[3ex] 2\;\;Democrats\;\;from\;\;5\;\;Democrats \\[3ex] Number\;\;of\;\;ways \\[3ex] = C(16, 12) * C(5, 2) \\[3ex] = \dfrac{16!}{(16 - 12)! * 12!} * \dfrac{5!}{(5 - 2)! * 2!} \\[5ex] = \dfrac{16!}{4! * 12!} * \dfrac{5!}{3! * 2!} \\[5ex] = \dfrac{16 * 15 * 14 * 13 * 12!}{4 * 3 * 2 * 1 * 12!} * \dfrac{5 * 4 * 3!}{3! * 2 * 1} \\[5ex] = 4 * 5 * 7 * 13 * 5 * 2 \\[3ex] = 18200\;\;ways \\[3ex] (d.) \\[3ex] We\;\;can\;\;solve\;\;this\;\;question\;\;in\;\;at\;\;least\;\;two\;\;ways \\[3ex] \underline{First\;\;Method} \\[3ex] Select \\[3ex] 12\;\;Republicans\;\;from\;\;21\;\;politicians \\[3ex] Remaining:\;\; 21 - 12 = 9\;\;politicians \\[3ex] Then\;\;select \\[3ex] 2\;\;Democrats\;\;from\;\;9\;\;politicians \\[3ex] Number\;\;of\;\;ways \\[3ex] = C(21, 12) * C(9, 2) \\[3ex] = 293930 * \dfrac{9!}{(9 - 2)! * 2!} \\[5ex] = 293930 * \dfrac{9!}{7! * 2!} \\[5ex] = 293930 * \dfrac{9 * 8 * 7!}{7! * 2 * 1} \\[5ex] = 293930 * 9 * 4 \\[3ex] = 10581480\;\;ways \\[3ex] \underline{Second\;\;Method} \\[3ex] Select \\[3ex] 2\;\;Democrats\;\;from\;\;21\;\;politicians \\[3ex] Remaining:\;\; 21 - 2 = 19\;\;politicians \\[3ex] Then\;\;select \\[3ex] 12\;\;Democrats\;\;from\;\;19\;\;politicians \\[3ex] Number\;\;of\;\;ways \\[3ex] = C(21, 2) * C(19, 12) \\[3ex] = \dfrac{21!}{(21 - 2)! * 2!} * \dfrac{19!}{(19 - 12)! * 12!} \\[5ex] = \dfrac{21!}{19! * 2!} * \dfrac{19!}{7! * 12!} \\[5ex] = \dfrac{21 * 20 * 19!}{19! * 2 * 1} * \dfrac{19 * 18 * 17 * 16 * 15 * 14 * 13 * 12!}{7 * 6 * 5 * 4 * 3 * 2 * 1 * 12!} \\[5ex] = 21 * 10 * 19 * 17 * 4 * 3 * 13 \\[3ex] = 10581480\;\;ways $

(18.) ACT Yvette has 6 pairs of leggings, 2 pairs of shoes, and 6 T-shirts, which all go together well.
How many different groupings of 1 of her 6 pairs of leggings, 1 of her 2 pairs of shoes, and 1 of her 6 T-shirts are available for Yvette to wear?

$ F.\;\; 8 \\[3ex] G.\;\; 12 \\[3ex] H.\;\; 14 \\[3ex] J.\;\; 24 \\[3ex] K.\;\; 72 \\[3ex] $ Similar to Questions (1.) and (3.)

First Approach: Fundamental Counting Principle

$ Number\;\;of\;\;ways \\[3ex] = 6 * 2 * 6 \\[3ex] = 72\;\;different\;\;groupings \\[5ex] $ Second Approach: Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] Number\;\;of\;\;ways \\[3ex] = C(6, 1) * C(2, 1) * C(6, 1) \\[3ex] = \dfrac{6!}{(6 - 1)! * 1!} * \dfrac{2!}{(2 - 1)! * 1!} * \dfrac{6!}{(6 - 1)! * 1!} \\[5ex] = \dfrac{6 * 5!}{5! * 1} * \dfrac{2 * 1!}{1! * 1} * \dfrac{6 * 5!}{5! * 1} \\[5ex] = 6 * 2 * 6 \\[3ex] = 72\;\;different\;\;groupings $

(20.) ACT In a window display at a flower shop, there are 3 spots for 1 plant each.
To fill these 3 spots, Emily has 6 plants to select from, each of a different type.
Selecting from the 6 plants, Emily can make how many possible display arrangements with 1 plant in each spot?
(Note: The positions of the unselected plants do not matter.)

$ A.\;\; 3 \\[3ex] B.\;\; 6 \\[3ex] C.\;\; 15 \\[3ex] D.\;\; 120 \\[3ex] E.\;\; 216 \\[3ex] $ Before you attempt this question, did you notice Question (14.)?

Fundamental Counting Principle and Permutation

Fundamental Counting Principle

$3$ spots for $1$ plant
Any $1$ of $6$ different plants can be selected to fill the first spot.
This means that the first spot can be filled by any one of the six plants.
This means that the first spot can be filled in six ways.

Once the first spot is filled by one plant, two other spots remain for the remaining five plants.
The second spot can be filled by any one of the five remaining plants.
This means that the second spot can be filled in five ways.

Once the second spot is filled, the third spot can be filled by any one of the remaining four plants.
This means that the third spot can be filled in four ways.

$ \underline{1st\:spot}\:\:\:\underline{2nd\:spot}\:\:\:\underline{3rd\:spot} \\[3ex] \;\;\;6 \:\:\:\:\:\:\;\;\; * \;\;\;\:\:\: 5 \;\;\;\:\:\: * \;\;\;\:\: 4 \\[3ex] = 120\;possible\;\;arrangements \\[3ex] $ The positions of the unselected plants do not matter.
However, the order does matter.
The order is the first, second, and third spots.
Any plant selected from the sevem plants can fill these spots.
However, the spots are ordered: first spot, second spot, third spot.
Therefore it is also Permutation

$ P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] Permutation\;\;of\;\;6\;plants\;\;taking\;\;3\;spots\;\;at\;\;a\;\;time \\[3ex] n = 6 \\[3ex] r = 3 \\[3ex] Number\;\;of\;\;arrangements \\[3ex] = P(6, 3) \\[3ex] = \dfrac{67!}{(6 - 3)!} \\[5ex] = \dfrac{6!}{3!} \\[5ex] = \dfrac{6 * 5 * 4 * 3!}{3!} \\[5ex] = 6 * 5 * 4 \\[3ex] = 120\;possible\;\;arrangements $

Top

(23.) JAMB How many possible ways are there of seating seven people P, Q, R, S, T, U and V at a circular table?

$ A.\;\; 360 \\[3ex] B.\;\; 720 \\[3ex] C.\;\; 2520 \\[3ex] D.\;\; 5040 \\[3ex] $

Circular Permutation
P, Q, R, S, T, U, V = 7 people

$ Number\;\;of\;\;circular\;\;permutations = (n - 1)! \\[3ex] = (7 - 1)! \\[3ex] = 6! \\[3ex] = 720\;\;ways $

(24.) HSC Mathematics Extension 1 A committee containing 5 men and 3 women is to be formed from a group of 10 men and 8 women.
In how many different ways can the committee be formed?

This is a case of Combination because there is no specific man woman
It can be any 5 men from the 10 men and any 3 women from the 8 women

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] Number\;\;of\;\;different\;\;ways \\[3ex] = C(10, 5) * C(8, 3) \\[3ex] = \dfrac{10!}{(10 - 5)! 5!} * \dfrac{8!}{(8 - 3)! 3!} \\[5ex] = \dfrac{10 * 9 * 8 * 7 * 6 * 5!}{5! * 5 * 4 * 3 * 2 * 1} * \dfrac{8 * 7 * 6 * 5!}{5! * 3 * 2 * 1} \\[5ex] = 2 * 9 * 2 * 7 * 8 * 7 \\[3ex] = 14112\;\;ways $

(26.) ACT Mark and Juanita own a sandwich shop.
They offer 3 kinds of bread, 5 kinds of meat, and 3 kinds of cheese.
Each type of sandwich has a combination of exactly 3 ingredients: 1 bread, 1 meat, and 1 cheese.
How many types of sandwiches are possible?

$ A.\;\; 11 \\[3ex] B.\;\; 15 \\[3ex] C.\;\; 30 \\[3ex] D.\;\; 45 \\[3ex] E.\;\; 120 \\[3ex] $ Similar to Questions (1.), (3.), and (18.)

First Approach: Fundamental Counting Principle

$ Number\:\:of\:\:types\:\:of\;\;samdwiches \\[3ex] = 3 * 5 * 3 \\[3ex] = 45\:sandwiches \\[5ex] $ Second Approach: Combination

Number of types of sandwiches = number of 3-ingredient combinations

$ Number\:\:of\:\:3-ingredient\:\:combinations \\[3ex] = C(3, 1) * C(5, 1) * C(3, 1) \\[3ex] C(n, r) = \dfrac{n!}{(n - r)!r!} \\[5ex] C(3, 1) \\[3ex] = \dfrac{3!}{(3 - 1)! * 1!} \\[5ex] = \dfrac{3 * 2!}{2! * 1} \\[5ex] = 3 \\[3ex] C(5, 1) \\[3ex] = \dfrac{5!}{(5 - 1)!1!} \\[5ex] = \dfrac{5 * 4!}{4! * 1} \\[5ex] = 5 \\[3ex] \implies \\[3ex] Number\:\:of\:\:types\:\:of\;\;samdwiches \\[3ex] = 3 * 5 * 3 = 45\:sandwiches $

(32.) ACT Kareem has 4 sweaters, 6 shorts, and 3 pairs of slacks.
How many distinct outfits, each consisting of a sweater, a shirt, and a pair of slacks, can Kareem select?

$ F.\;\; 13 \\[3ex] G.\;\; 36 \\[3ex] H.\;\; 42 \\[3ex] J.\;\; 72 \\[3ex] K.\;\; 216 \\[3ex] $ Similar to Questions (1.), (3.), (18.), and (26.)

First Approach: Fundamental Counting Principle

$ Number\:\:of\:\:distinct\;\;outfits \\[3ex] = 4 * 6 * 3 \\[3ex] = 72\:outfits \\[5ex] $ Second Approach: Combination

$ Number\:\:of\:\:distinct\;\;outfits \\[3ex] = C(4, 1) * C(6, 1) * C(3, 1) \\[3ex] C(n, r) = \dfrac{n!}{(n - r)!r!} \\[5ex] C(4, 1) \\[3ex] = \dfrac{4!}{(4 - 1)! * 1!} \\[5ex] = \dfrac{4 * 3!}{3! * 1} \\[5ex] = 4 \\[3ex] C(6, 1) \\[3ex] = \dfrac{6!}{(6 - 1)!1!} \\[5ex] = \dfrac{6 * 5!}{5! * 1} \\[5ex] = 6 \\[3ex] \implies \\[3ex] Number\:\:of\:\:distinct\;\;outfits \\[3ex] \\[3ex] = 4 * 6 * 3 \\[3ex] = 72\:outfits $