# 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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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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