# Permutation and Combination Questions Answers

Permutation and Combination Questions Answers (MCQ) listings with explanations includes questions on calculating combinations, permutations by different tricks which are important for many competitive exams.

1) Evaluate \begin{aligned} \frac{30!}{28!} \end{aligned}

• 1) 970
• 2) 870
• 3) 770
• 4) 670
Ans.   B
Explanation :
\begin{aligned} = \frac{30!}{28!} \ = \frac{30 * 29 * 28!}{28!} \ = 30 * 29 = 870 \end{aligned}

2) Evaluate permutation equation \begin{aligned} ^{59}{P}_3 \end{aligned}

• 1) 195052
• 2) 195053
• 3) 195054
• 4) 185054
Ans.   C
Explanation :
\begin{aligned} ^n{P}_r = \frac{n!}{(n-r)!} \^{59}{P}_3 = \frac{59!}{(56)!} \= \frac{59 * 58 * 57 * 56!}{(56)!} \= 195054 \end{aligned}

3) Evaluate permutation \begin{aligned} ^5{P}_5 \end{aligned}

• 1) 120
• 2) 110
• 3) 98
• 4) 24
Ans.   A
Explanation :
\begin{aligned} ^n{P}_n = n! \^5{P}_5 = 5*4*3*2*1 \= 120 \end{aligned}

4) Evaluate permutation equation \begin{aligned} ^{75}{P}_2\end{aligned}

• 1) 5200
• 2) 5300
• 3) 5450
• 4) 5550
Ans.   D
Explanation :
\begin{aligned} ^n{P}_r = \frac{n!}{(n-r)!} \^{75}{P}_2 = \frac{75!}{(75-2)!} \= \frac{75*74*73!}{(73)!} \= 5550 \end{aligned}

5) Evaluate combination \begin{aligned} ^{100}{C}_{97} = \frac{100!}{(97)!(3)!} \\end{aligned}

• 1) 161700
• 2) 151700
• 3) 141700
• 4) 131700
Ans.   A
Explanation :
\begin{aligned} ^{n}{C}_r = \frac{n!}{(r)!(n-r)!} \^{100}{C}_{97} = \frac{100!}{(97)!(3)!} \= \frac{100*99*98*97!}{(97)!(3)!} \= \frac{100*99*98}{3*2*1} \= \frac{100*99*98}{3*2*1} \= 161700 \end{aligned}

6) Evaluate combination \begin{aligned} ^{100}{C}_{100} \end{aligned}

• 1) 10000
• 2) 1000
• 3) 10
• 4) 1
Ans.   D
Explanation :
\begin{aligned} ^{n}{C}_{n} = 1 \ ^{100}{C}_{100} = 1 \end{aligned}

7) How many words can be formed by using all letters of TIH

• 1) 100
• 2) 120
• 3) 140
• 4) 160
Ans.   B
Explanation :
First thing to understand in this question is that it is a permutation question. Total number of words = 5 Required number = \begin{aligned} ^5{P}_5 = 5! \= 5*4*3*2*1 = 120 \end{aligned}

8) In how many words can be formed by using all letters of the word BHOP

• 1) 420
• 2) 520
• 3) 620
• 4) 720
Ans.   D
Explanation :
Required number \begin{aligned} = 6! \ = 6*5*4*3*2*1 \ = 720 \end{aligned}

9) In how many way the letter of the word 'APPLE' can be arrang

• 1) 20
• 2) 40
• 3) 60
• 4) 80
Ans.   C
Explanation :
Friends the main point to note in this question is letter 'P' is written twice in the word. Easy way to solve this type of permutation question is as, So word APPLE contains 1A, 2P, 1L and 1E Required number = \begin{aligned} = \frac{5!}{1!*2!*1!*1!} \ = \frac{5*4*3*2!}{2!} \ = 60 \end{aligned}

10) In how many ways can the letters of the CHEATER be arran

• 1) 20160
• 2) 2520
• 3) 360
• 4) 80
Ans.   B
Explanation :
As we can see the letter 'E' is twice in given word, so Required Number \begin{aligned} = \frac{7!}{2!} \\ = \frac{7*6*5*4*3*2!}{2!} \\ = 2520 \end{aligned}

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