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It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it. - Samuel Dominic Chukwuemeka

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

Solved Examples 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

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.) Adv Maths Exams - P1 The value of 6 factorial (6!) is:

$ A.\;\;36 \;\;\;\;\;\;\;\; B.\;\; 720 \\[3ex] C.\;\;6 \;\;\;\;\;\;\;\;\; D.\;\; 6^6 \\[3ex] $

Factorial Notation

$ 6! \\[3ex] = 6 * 5 * 4 * 3 * 2 * 1 \\[3ex] = 720 $

(2.) ACT The positive integer $n!$ is defined as the product of all the positive integers less than or equal to $n$.
For example, $3! = 1(2)(3) = 6$.
What is the value of the expression $\dfrac{6!}{3!2!}$?

Factorial Notation

$ \dfrac{6!}{3!2!} \\[5ex] = \dfrac{6 * 5 * 4 * 3!}{3! * 2 * 1} \\[5ex] = 6 * 5 * 2 \\[3ex] = 60 $

(3.) $\dfrac{9!}{3!4!2!}$

Factorial Notation

$ \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 $

(4.) Adv Maths Exams - P1 The number of permutations of 5 objects is:

$ A.\;\;25 \;\;\;\;\;\;\;\; B.\;\; 5 \\[3ex] C.\;\;5^5 \;\;\;\;\;\;\;\; D.\;\; 120 \\[3ex] $

Factorial Notation

$ 5! \\[3ex] = 5 * 4 * 3 * 2 * 1 \\[3ex] = 120 \\[3ex] $ OR

Permutation

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

(5.) $P(7, 4)$

Permutation

$ P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] P(7, 4) \\[3ex] = \dfrac{7!}{(7 - 4)!} \\[5ex] = \dfrac{7!}{3!} \\[5ex] = \dfrac{7 * 6 * 5 * 4 * 3!}{3!} \\[5ex] = 7 * 6 * 5 * 4 \\[3ex] = 840 $

(6.) $P(30, 3)$

Permutation

$ P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] P(30, 3) \\[3ex] = \dfrac{30!}{(30 - 3)!} \\[5ex] = \dfrac{30!}{27!} \\[5ex] = \dfrac{30 * 29 * 28 * 27!}{27!} \\[5ex] = 30 * 29 * 28 \\[3ex] = 24360 $

(7.) $C(21, 12)$

(8.) $C(30, 3)$

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] 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] = 4060 $

(9.) $C(21, 12) * C(9, 2)$

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] 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 * 5 * 14 * 13}{1 * 2} \\[5ex] = 19 * 17 * 5 * 14 * 13 \\[3ex] = 293930 \\[5ex] C(9, 2) \\[3ex] = \dfrac{9!}{(9 - 2)! * 2!} \\[5ex] = \dfrac{9!}{7! * 2!} \\[5ex] = \dfrac{9 * 8 * 7!}{7! * 2 * 1} \\[5ex] = 9 * 4 \\[3ex] = 36 \\[5ex] C(21, 12) * C(9, 2) \\[3ex] = 293930 * 36 \\[3ex] = 10581480 $

(10.) $C(16, 12) * C(5, 2)$

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] C(16, 12) \\[3ex] = \dfrac{16!}{(16 - 12)! * 12!} \\[5ex] = \dfrac{16!}{4! * 12!} \\[5ex] = \dfrac{16 * 15 * 14 * 13 * 12!}{4 * 3 * 2 * 1 * 12!} \\[5ex] = 4 * 5 * 7 * 13 \\[3ex] = 1820 \\[5ex] 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 \\[5ex] C(16, 12) * C(5, 2) \\[3ex] = 1820 * 10 \\[3ex] = 18200 $

(11.) $C(21, 2) * C(19, 12)$

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] C(21, 2) \\[3ex] = \dfrac{21!}{(21 - 2)! * 2!} \\[5ex] = \dfrac{21!}{19! * 2!} \\[5ex] = \dfrac{21 * 20 * 19!}{19! * 2 * 1} \\[5ex] = 21 * 10 \\[3ex] = 210 \\[5ex] C(19, 12) \\[3ex] = \dfrac{19!}{(19 - 12)! * 12!} \\[5ex] = \dfrac{19!}{7! * 12!} \\[5ex] = \dfrac{19 * 18 * 17 * 16 * 15 * 14 * 13 * 12!}{7 * 6 * 5 * 4 * 3 * 2 * 1 * 12!} \\[5ex] = 19 * 17 * 4 * 3 * 13 \\[3ex] = 50388 \\[5ex] C(21, 2) * C(19, 12) \\[3ex] = 210 * 50388 \\[3ex] = 10581480 $

(12.) $C(5, 2) * C(15, 8)$

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] 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 \\[5ex] C(15, 8) \\[3ex] = \dfrac{15!}{(15 - 8)! * 8!} \\[5ex] = \dfrac{15!}{7! * 8!} \\[5ex] = \dfrac{15 * 14 * 13 * 12 * 11 * 10 * 9 * 8!}{7 * 6 * 5 * 4 3 * 2 * 1 * 8!} \\[5ex] = \dfrac{13 * 2 * 11 * 10 * 9}{4} \\[5ex] = \dfrac{13 * 11 * 10 * 9}{2} \\[5ex] = 13 * 11 * 5 * 9 \\[3ex] = 6435 \\[5ex] C(5, 2) * C(15, 8) \\[3ex] = 10 * 6435 \\[3ex] = 64350 $

(13.) $\dfrac{15!}{7!6!2!}$

Factorial Notation

$ \dfrac{15!}{7!6!2!} \\[5ex] = \dfrac{15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7!}{7! * 6 * 5 * 4 * 3 * 2 * 1 * 2 * 1} \\[5ex] = 14 * 13 * 11 * 10 * 9 \\[3ex] = 180180 $

(14.) $\dfrac{12!}{4!4!4!}$

Factorial Notation

$ \dfrac{12!}{4!4!4!} \\[5ex] = \dfrac{12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4!}{4! * 4 * 3 * 2 * 1 * 4 * 3 * 2 * 1} \\[5ex] = 11 * 10 * 9 * 7 * 5 \\[3ex] = 34650 $

(15.) $C(4, 3) * C(4, 2) * C(9, 5) * [C(5, 1) + C(3, 1)]$

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] C(4, 3) * C(4, 2) * C(9, 5) \\[3ex] = \dfrac{4!}{(4 - 3)! * 3!} * \dfrac{4!}{(4 - 2)! * 2!} * \dfrac{9!}{(9 - 5)! * 5!} \\[5ex] = \dfrac{4!}{1! * 3!} * \dfrac{4!}{2! * 2!} * \dfrac{9!}{4! * 5!} \\[5ex] = \dfrac{1}{1 * 3!} * \dfrac{4 * 3!}{2 * 1 * 2 * 1} * \dfrac{9 * 8 * 7 * 6 * 5!}{1 * 5!} \\[5ex] = 9 * 8 * 7 * 6 \\[3ex] = 3024 \\[5ex] C(5, 1) \\[3ex] = \dfrac{5!}{(5 - 1)! * 1!} \\[5ex] = \dfrac{5 * 4!}{4! * 1!} \\[5ex] = 5 \\[5ex] C(3, 1) \\[3ex] = \dfrac{3!}{(3 - 1)! * 1!} \\[5ex] = \dfrac{3 * 2!}{2! * 1!} \\[5ex] = 3 \\[5ex] C(4, 3) * C(4, 2) * C(9, 5) * [C(5, 1) + C(3, 1)] \\[3ex] = 3024 * [5 + 3] \\[3ex] = 3024 * 8 \\[3ex] = 24192 $

(16.) $P(n, 1)$

Permutation

$ P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] P(n, 1) \\[3ex] = \dfrac{n!}{(n - 1)!} \\[5ex] = \dfrac{n * (n - 1)!}{(n - 1)!} \\[5ex] = n $

(17.)

$ [C(6, 1) * C(9, 3)] + \\[3ex] [C(6, 2) * C(9, 2)] + \\[3ex] [C(6, 3) * C(9, 1)] + \\[3ex] [C(6, 4) * C(9, 0)] \\[3ex] $

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] C(6, 1) * C(9, 3) \\[5ex] = \dfrac{6!}{(6 - 1)! 1!} * \dfrac{9!}{(9 - 3)! 3!} \\[5ex] = \dfrac{6!}{5! 1!} * \dfrac{9!}{6! 3!} \\[5ex] = \dfrac{6 * 5!}{5!} * \dfrac{9 * 8 * 7 * 6!}{6! * 3 * 2 * 1} \\[5ex] = 9 * 8 * 7 \\[3ex] = 504 \\[5ex] C(6, 2) * C(9, 2) \\[5ex] = \dfrac{6!}{(6 - 2)! 2!} * \dfrac{9!}{(9 - 2)! 2!} \\[5ex] = \dfrac{6!}{4! 2!} * \dfrac{9!}{7! 2!} \\[5ex] = \dfrac{6 * 5 * 4!}{4! * 2 * 1} * \dfrac{9 * 8 * 7!}{7! * 2 * 1} \\[5ex] = 6 * 5 * 9 * 2 \\[3ex] = 540 \\[5ex] C(6, 3) * C(9, 1) \\[5ex] = \dfrac{6!}{(6 - 3)! 3!} * \dfrac{9!}{(9 - 1)! 1!} \\[5ex] = \dfrac{6!}{3! 3!} * \dfrac{9!}{8! 2!} \\[5ex] = \dfrac{6 * 5 * 4 * 3!}{3! * 3 * 2 * 1} * \dfrac{9 * 8!}{8! * 1!} \\[5ex] = 5 * 4 * 9 \\[3ex] = 180 \\[5ex] C(6, 4) * C(9, 0) \\[5ex] = \dfrac{6!}{(6 - 4)! 4!} * \dfrac{9!}{(9 - 0)! 0!} \\[5ex] = \dfrac{6!}{2! 4!} * \dfrac{9!}{9! 0!} \\[5ex] = \dfrac{6 * 5 * 4!}{2 * 1 * 4!} * \dfrac{1}{1} \\[5ex] = 3 * 5 * 1 \\[3ex] = 15 \\[5ex] \therefore [C(6, 1) * C(9, 3)] + [C(6, 2) * C(9, 2)] + [C(6, 3) * C(9, 1)] + [C(6, 4) * C(9, 0)] \\[3ex] = 504 + 540 + 180 + 15 \\[3ex] = 1239 $

(18.) JAMB If $^6P_r = 6$, find the value of $^6P_{r + 1}$

$ A.\;\; 30 \\[3ex] B.\;\; 33 \\[3ex] C.\;\; 35 \\[3ex] D.\;\; 15 \\[3ex] $

Permutation

$ ^nP_r = \dfrac{n!}{(n - r)!} \\[5ex] ^6P_r = 6 \\[3ex] ^6P_1 = 6 \\[3ex] \implies r = 1 \\[3ex] r + 1 = 1 + 1 = 2 \\[3ex] ^6P_{r + 1} = ^6P_{2} \\[3ex] = \dfrac{6!}{(6 - 2)!} \\[5ex] = \dfrac{6!}{4!} \\[5ex] = \dfrac{6 * 5 * 4!}{4!} \\[5ex] = 6 * 5 \\[3ex] = 30 $

(19.) $C(20, 18) + C(20, 19) + C(20, 20)$

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] C(20, 18) \\[3ex] = \dfrac{20!}{(20 - 18)! 18!} \\[5ex] = \dfrac{20!}{2! * 18!} \\[5ex] = \dfrac{20 * 19 * 18!}{2 * 1 * 18!} \\[5ex] = 10 * 19 \\[3ex] = 190 \\[5ex] C(20, 19) \\[3ex] = \dfrac{20!}{(20 - 19)! 19!} \\[5ex] = \dfrac{20!}{1! * 19!} \\[5ex] = \dfrac{20 * 19!}{1 * 19!} \\[5ex] = 20 \\[5ex] C(20, 20) \\[3ex] = \dfrac{20!}{(20 - 20)! 20!} \\[5ex] = \dfrac{20!}{0! * 20!} \\[5ex] = \dfrac{20!}{1 * 20!} \\[5ex] = 1 \\[5ex] C(20, 18) + C(20, 19) + C(20, 20) \\[3ex] = 190 + 20 + 1 \\[3ex] = 211 $

(20.) $C(4, r)$

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] C(4, r) \\[3ex] = \dfrac{4!}{(4 - r)! r!} \\[5ex] = \dfrac{4 * 3 * 2 * 1}{(4 - r)! r!} \\[5ex] = \dfrac{24}{(4 - r)! r!} \\[5ex] $

Top

(21.) $C(n, n - 2)$

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] C(n, n - 2) \\[3ex] = \dfrac{n!}{[n - (n - 2)]! * (n - 2)!} \\[5ex] = \dfrac{n!}{(n - n + 2)! * (n - 2)!} \\[5ex] = \dfrac{n!}{2! * (n - 2)!} \\[5ex] = \dfrac{n * (n - 1) * (n - 2)!}{2 * 1 * (n - 2)!} \\[5ex] = \dfrac{n(n - 1)}{2} $

(22.) $P(n, n - 2)$

Permutation

$ P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] P(n, n - 2) \\[3ex] = \dfrac{n!}{[n - (n - 2)]!} \\[5ex] = \dfrac{n!}{(n - n + 2)!} \\[5ex] = \dfrac{n!}{2!} \\[5ex] = \dfrac{n!}{2 * 1} \\[5ex] = \dfrac{n!}{2} $

(23.) $\dfrac{10!}{2!2!2!}$

Factorial Notation

$ \dfrac{10!}{2!2!2!} \\[5ex] = \dfrac{10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2!}{2! * 2 * 1 * 2 * 1} \\[5ex] = 10 * 9 * 8 * 7 * 6 * 5 * 3 \\[3ex] = 453600 $

(24.) $C(r, 4)$

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] C(r, 4) \\[3ex] = \dfrac{r!}{(r - 4)! 4!} \\[5ex] = \dfrac{r * (r - 1) * (r - 2) * (r - 3) * (r - 4)!}{(r - 4)! * 4 * 3 * 2 * 1} \\[5ex] = \dfrac{r(r - 1)(r - 2)(r - 3)}{24} $

(25.) $\dfrac{9!}{3!2!2!}$

Factorial Notation

$ \dfrac{9!}{3!2!2!} \\[5ex] = \dfrac{9 * 8 * 7 * 6 * 5 * 4 * 3!}{3! * 2 * 1 * 2 * 1} \\[5ex] = 9 * 8 * 7 * 6 * 5 \\[3ex] = 15120 $

(26.) $P(k, 2)$

Permutation

$ P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] P(k, 2) \\[3ex] = \dfrac{k!}{(k - 2)!} \\[5ex] = \dfrac{k * (k - 1) * (k -2)!}{(k - 2)!} \\[5ex] = k(k - 1) $

(28.) Adv Maths Exams - P1 How many combinations of 5 objects from 7 are there?

Combination

$ C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] C(7, 5) \\[3ex] = \dfrac{7!}{(7 - 5)! 5!} \\[5ex] = \dfrac{7 * 6 * 5!}{2! * 5!} \\[5ex] = \dfrac{7 * 6}{2 * 1} \\[5ex] = 7 * 3 \\[3ex] = 21 $

(29.) WASSCE-FM Given that $\displaystyle{\binom{n}{r}} = ^nC_r$,

simplify $\displaystyle{\binom{2x + 1}{3}} - \displaystyle{\binom{2x - 1}{3}} - 2\displaystyle{\binom{x}{2}}$

$ \displaystyle{\binom{n}{r}} = \dfrac{n!}{(n - r)! r!} \\[5ex] n! = n * (n - 1) * (n - 2)!...among\;\;others \\[3ex] \displaystyle{\binom{2x + 1}{3}} - \displaystyle{\binom{2x - 1}{3}} - 2\displaystyle{\binom{x}{2}} \\[5ex] \underline{First\;\;Part} \\[3ex] \displaystyle{\binom{2x + 1}{3}} \\[3ex] = \dfrac{(2x + 1)!}{(2x + 1 - 3)! * 3!} \\[5ex] = \dfrac{(2x + 1)!}{(2x - 2)! * 3!} \\[5ex] = \dfrac{(2x + 1) * 2x * (2x - 1) * (2x - 2)!}{(2x - 2)! * 3 * 2 * 1} \\[5ex] = \dfrac{x(2x + 1)(2x - 1)}{3} \\[7ex] \underline{Second\;\;Part} \\[3ex] \displaystyle{\binom{2x - 1}{3}} \\[3ex] = \dfrac{(2x - 1)!}{(2x - 1 - 3)! * 3!} \\[5ex] = \dfrac{(2x - 1)!}{(2x - 4)! * 3!} \\[5ex] = \dfrac{(2x - 1) * (2x - 2) * (2x - 3) * (2x - 4)!}{(2x - 4)! * 3 * 2 * 1} \\[5ex] = \dfrac{(2x - 1) * 2(x - 1) * (2x - 3)}{6} \\[5ex] = \dfrac{(2x - 1)(x - 1)(2x - 3)}{3} \\[7ex] \underline{Third\;\;Part} \\[3ex] 2\displaystyle{\binom{x}{2}} \\[5ex] = 2 * \dfrac{x!}{(x - 2)! 2!} \\[5ex] = 2 * \dfrac{x * (x - 1) * (x - 2)!}{(x - 2)! * 2 * 1} \\[5ex] = \dfrac{2 * x * (x - 1)}{2} \\[5ex] = x(x - 1) \\[5ex] \implies \\[3ex] \displaystyle{\binom{2x + 1}{3}} - \displaystyle{\binom{2x - 1}{3}} - 2\displaystyle{\binom{x}{2}} \\[5ex] = \dfrac{x(2x + 1)(2x - 1)}{3} - \dfrac{(2x - 1)(x - 1)(2x - 3)}{3} - x(x - 1) \\[5ex] = \dfrac{x(2x + 1)(2x - 1)}{3} - \dfrac{(2x - 1)(x - 1)(2x - 3)}{3} - \dfrac{x(x - 1)}{1} \\[5ex] = \dfrac{[x(2x + 1)(2x - 1)] - [(2x - 1)(x - 1)(2x - 3)] - [3x(x - 1)]}{3} \\[5ex] = \dfrac{x[(2x)^2 - 1^2] - [(2x^2 - 2x - x + 1)(2x - 3)] - (3x^2 - 3x)}{3} \\[5ex] = \dfrac{x(4x^2 - 1) - [(2x^2 - 3x + 1)(2x - 3)] - 3x^2 + 3x}{3} \\[5ex] = \dfrac{4x^3 - x - (4x^3 - 6x^2 - 6x^2 + 9x + 2x - 3) - 3x^2 + 3x}{3} \\[5ex] = \dfrac{4x^3 - x - (4x^3 - 12x^2 + 11x - 3) - 3x^2 + 3x}{3} \\[5ex] = \dfrac{4x^3 - x - 4x^3 + 12x^2 - 11x + 3 - 3x^2 + 3x}{3} \\[5ex] = \dfrac{9x^2 - 9x + 3}{3} \\[5ex] = \dfrac{3(3x^2 - 3x + 1)}{3} \\[5ex] = 3x^2 - 3x + 1 $

(30.) JAMB If $\dfrac{^6C_r}{^6P_r} = \dfrac{1}{6}$, find the value of $r$

$ A.\;\; 1 \\[3ex] B.\;\; 3 \\[3ex] C.\;\; 5 \\[3ex] D.\;\; 6 \\[3ex] $

Combination and Permutation

$ \dfrac{^6C_r}{^6P_r} = \dfrac{1}{6} \\[5ex] But:\;\; ^6P_r = r! \;*\; ^6C_r \\[3ex] Because:\;\; ^nP_r = r! \;*\; ^nC_r \\[3ex] \implies \\[3ex] \dfrac{^6C_r}{r! \;*\; ^6C_r} = \dfrac{1}{6} \\[5ex] \dfrac{1}{r!} = \dfrac{1}{6} \\[5ex] r! = 6 \\[3ex] r = 3 $

(31.) ACT For all positive integers n, which of the following is a correct ordering of the terms $n^n,\;\;\;(n!)^n,\;\;and\;\;(n!)^{n!}$?
(Note: $n! = (n)(n - 1)(n - 2) ... (2)(1)$)

$ A.\;\; n^n \ge (n!)^n \ge (n!)^{n!} \\[3ex] B.\;\; (n!)^n \ge n^n \ge (n!)^{n!} \\[3ex] C.\;\; (n!)^n \ge (n!)^{n!} \ge n^n \\[3ex] D.\;\; (n!)^{n!} \ge (n!)^n \ge n^n \\[3ex] E.\;\; (n!)^{n!} \ge n^n \ge (n!)^n \\[3ex] $

Based on the definition:
$n! \ge n$ because $n!$ covers the product of $n$ and othe positive integers less than $n$
We use ≥ because for the integers 1 and 2; $n!$ is equal to $n$
$1! = 1$
$2! = 2 * 1 = 2$
But for all other positive integers greater than 2, $n! \gt n$
$3! \gt 3$ because $3! = 3(2)(1) = 6$ and $6 \gt 3$
$4! \gt 4$ because $4! = 4(3)(2)(1) = 24$ and $24 \gt 4$
So, generally ; for all positive integers:
$n! \ge n$

Similarly:
For the exponents (which are also positive integers):
For the positive integers 1 and 2; $n!^{n!}$ is equal to $n^n$

$ 1!^{1!} = 1^1 = 1 \\[3ex] 2!^{2!} = 2^2 = 4 \\[3ex] $ For all other positive integers (positive integers greater than 2):
$n!^{n!}$ is greater than $n^n$

$ 3!^{3!} \gt 3^3 \\[3ex] 6^6 \gt 3^3 \\[3ex] Also: \\[3ex] 3!^{3!} \gt (3!)^3 \gt 3^3 \\[3ex] 6^6 \gt 6^3 \gt 3^3 \\[3ex] $ All in all (considering the integers 1 and 2, as well as other positive integers greater than 2):

$ (n!)^{n!} \ge (n!)^n \ge n^n $

(32.) ACT What is the value of the expression $\dfrac{8!}{(4!)^2}$ ?

(Note: 3! = 3(2)(1) and 6! = 6(5)(4)(3)(2)(1))

$ F.\;\; 0 \\[3ex] G.\;\; \dfrac{1}{2} \\[5ex] H.\;\; 1 \\[3ex] J.\;\; 70 \\[3ex] K.\;\; 420 \\[3ex] $

Factorial Notation

$ \dfrac{8!}{(4!)^2} \\[5ex] = \dfrac{8 * 7 * 6 * 5 * 4!}{4! * 4!} \\[5ex] = \dfrac{8 * 7 * 6 * 5}{4 * 3 * 2 * 1} \\[5ex] = 7 * 2 * 5 \\[3ex] = 70 $

(39.) JAMB If $^nP_3 - 6\left(^nC_4\right) = 0$, find the value of $n$

$ A.\;\; 6 \\[3ex] B.\;\; 5 \\[3ex] C.\;\; 8 \\[3ex] D.\;\; 7 \\[3ex] $

Permutation and Combination

$ r! \;*\; ^nC_r = ^nP_r \\[3ex] 3! \;*\; ^nC_3 = ^nP_3 \\[3ex] ^nP_r = \dfrac{n!}{(n - r)!} \\[5ex] ^nC_r = \dfrac{n!}{(n - r)! r!} \\[5ex] ^nC_3 = \dfrac{n!}{(n - 3)! 3!} \\[5ex] ^nC_4 = \dfrac{n!}{(n - 4)! 4!} \\[5ex] n! = n * (n - 1) * (n - 2)! \\[3ex] (n - 3)! = (n - 3) * (n - 4)! \\[3ex] 4! = 4 * 3 * 2 * 1 = 24 \\[3ex] ^nP_3 - 6\left(^nC_4\right) = 0 \\[5ex] \implies \\[3ex] 3! \;*\; ^nC_3 - 6\left(^nC_4\right) = 0 \\[5ex] 3! * \dfrac{n!}{(n - 3)! * 3!} - 6\left[\dfrac{n!}{(n - 4)! * 4!}\right] = 0 \\[5ex] \dfrac{n!}{(n - 4)!}\left[\dfrac{1}{n - 3} - \dfrac{6}{4!}\right] = 0 \\[5ex] Divide\;\;both\;\;sides\;\;by\;\; \dfrac{n!}{(n - 4)!} \\[5ex] \dfrac{1}{n - 3} - \dfrac{6}{24} = 0 \\[5ex] \dfrac{1}{n - 3} - \dfrac{1}{4} = 0 \\[5ex] \dfrac{1}{n - 3} = \dfrac{1}{4} \\[5ex] n - 3 = 4 \\[3ex] n = 4 + 3 \\[3ex] n = 7 \\[5ex] \underline{Check} \\[3ex] $

LHS	RHS
$ ^nP_3 - 6\left(^nC_4\right) \\[5ex] ^7P_3 - 6\left(^7C_4\right) \\[5ex] \dfrac{7!}{(7 - 3)!} - 6 * \dfrac{7!}{(7 - 4)! * 4!} \\[5ex] \dfrac{7 * 6 * 5 * 4!}{4!} - 6 * \left(\dfrac{7 * 6 * 5 * 4!}{3! * 4!}\right) \\[5ex] (7 * 6 * 5) - 6 * \left(\dfrac{7 * 6 * 5}{3 * 2 * 1}\right) \\[5ex] (7 * 6 * 5) - (6 * 7 * 5) \\[3ex] 0 $	$0$

(40.) ACT If $\dfrac{n!}{(n - 2)!} = 30$, then (n - 1)! = ?

$ A.\;\; 15 \\[3ex] B.\;\; 24 \\[3ex] C.\;\; 60 \\[3ex] D.\;\; 120 \\[3ex] E.\;\; 720 \\[3ex] $

Factorial Notation

$ n! = n(n - 1)(n - 2)! \\[3ex] \dfrac{n!}{(n - 2)!} = 30 \\[5ex] \dfrac{n(n - 1)(n - 2)!}{(n - 2)!} = 30 \\[5ex] n(n - 1) = 30 \\[3ex] n^2 - n - 30 = 0 \\[3ex] (n + 5)(n - 6) = 0 \\[3ex] n + 5 = 0 \;\;OR\;\; n - 6 = 0 \\[3ex] n = -5 \;\;OR\;\; n = 6 \\[3ex] n = -5\;(No,\;\;because\;\;n\;\;cannot\;\;be\;\;negative) \\[3ex] \therefore n = 6 \\[3ex] (n - 1)! \\[3ex] = (6 - 1)! \\[3ex] = 5! \\[3ex] = 5 * 4 * 3 * 2 * 1 \\[3ex] = 120 $

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(41.) The expression n! (read as n factorial) is defined as the product of all positive integers up to and including n, whenever n is a positive integer.
For example, $4! = 1 \cdot 2 \cdot 3 \cdot 4$
Whenever n is a positive integer, which of the following is equivalent to $\dfrac{(n + 1)!6!}{n!3!}$ ?

$ A.\;\; 120(n + 1) \\[3ex] B.\;\; 120 \\[3ex] C.\;\; 2 \\[3ex] D.\;\; \dfrac{2(n + 1)}{n} \\[5ex] E.\;\; \dfrac{(6n + 6)!}{(3n)!} \\[5ex] $

$ (n + 1)! = (n + 1) \cdot n! \\[3ex] 6! = 6 \cdot 5 \cdot 4 \cdot 3! \\[3ex] \dfrac{(n + 1)!6!}{n!3!} \\[5ex] = \dfrac{(n + 1) \cdot n! \cdot 6 \cdot 5 \cdot 4 \cdot 3!}{n! \cdot 3!} \\[5ex] = (n + 1) \cdot 120 \\[3ex] = 120(n + 1) $

(42.) ACT If k is an integer and k > 2, which of the following expressions is equivalent to $\dfrac{(k + 1)!(k - 1)!}{(k!)^2}$?
(Note: For any positive integer n.)
n! = (n)(n - 1)(n - 2) ... (2)(1). Hence
5! = (5)(4)(3)(2)(1) = 120.)

$ A.\;\; \dfrac{k + 1}{k} \\[5ex] B.\;\; k^2 - 1 \\[3ex] C.\;\; -1 \\[3ex] D.\;\; 0 \\[3ex] E.\;\; 1 \\[3ex] $

Factorial Notation

$ (k + 1)! = (k + 1) * k! \\[3ex] k! = k * (k - 1)! \\[3ex] \dfrac{(k + 1)!(k - 1)!}{(k!)^2} \\[5ex] = \dfrac{(k + 1) * k! * (k - 1)!}{k! * k!} \\[5ex] = \dfrac{(k + 1) * (k - 1)!}{k * (k - 1)!} \\[5ex] = \dfrac{k + 1}{k} $