CA Foundation P3 · Chapter 5

Basic Concepts of Permutations and Combinations MCQs with Answers

9 multiple-choice questions on Basic Concepts of Permutations and Combinations for CA Foundation P3 Quantitative Aptitude. Try each one before revealing the answer and explanation.

1. Question 1

   The number of different arrangements of the letters of the word LEDGER is:

   • A) 720
   • B) 180
   • C) 360
   • D) 120
   Show answer & explanation

   Answer: C) 360

   LEDGER has 6 letters with E repeated twice. Number of arrangements = 6!/2! = 720/2 = 360. Ignoring the repetition gives 720.

2. Question 2

   In how many ways can a committee of 3 be chosen from 8 persons?

   • A) 336
   • B) 24
   • C) 512
   • D) 56
   Show answer & explanation

   Answer: D) 56

   Order does not matter in a committee, so we use combinations: ⁸C₃ = 8 x 7 x 6 / (3 x 2 x 1) = 56. ⁸P₃ = 336 counts ordered selections, which is the usual error.

3. Question 3

   A committee of 5 is to be formed from 6 men and 4 women. In how many ways can this be done if the committee must contain exactly 2 women?

   • A) 252
   • B) 120
   • C) 60
   • D) 186
   Show answer & explanation

   Answer: B) 120

   Choose 2 women from 4: ⁴C₂ = 6. Choose 3 men from 6: ⁶C₃ = 20. Total = 6 x 20 = 120. ¹⁰C₅ = 252 ignores the condition, and 186 is the count for 'at least 2 women' (252 − 6 − 60).

4. Question 4

   In how many ways can 7 persons be seated around a circular table?

   • A) 720
   • B) 5040
   • C) 360
   • D) 2520
   Show answer & explanation

   Answer: A) 720

   For n persons around a circle the number of arrangements is (n − 1)!. Here (7 − 1)! = 6! = 720. 7! = 5,040 treats it as a row, and 360 = 6!/2 applies only to necklaces or garlands where clockwise and anticlockwise are the same.

5. Question 5

   How many 4-digit even numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7 if no digit is repeated?

   • A) 840
   • B) 480
   • C) 420
   • D) 360
   Show answer & explanation

   Answer: D) 360

   The units place must be even: 2, 4 or 6, giving 3 choices. The remaining three places are filled from the other 6 digits in ⁶P₃ = 6 x 5 x 4 = 120 ways. Total = 3 x 120 = 360. ⁷P₄ = 840 counts all 4-digit numbers, and 480 counts the odd ones.

6. Question 6

   If ⁿP₂ = 90, then n is:

   • A) 9
   • B) 45
   • C) 10
   • D) 15
   Show answer & explanation

   Answer: C) 10

   ⁿP₂ = n(n − 1) = 90. Since 10 x 9 = 90, n = 10 (the negative root −9 is rejected).

7. Question 7

   The number of diagonals of a polygon with 12 sides is:

   • A) 66
   • B) 54
   • C) 132
   • D) 42
   Show answer & explanation

   Answer: B) 54

   Lines joining any two vertices = ¹²C₂ = 66. Of these, 12 are sides. Diagonals = 66 − 12 = 54. Equivalently n(n − 3)/2 = 12 x 9 / 2 = 54.

8. Question 8

   In how many ways can the letters of the word ARTICLE be arranged so that the vowels occupy only the even positions?

   • A) 144
   • B) 720
   • C) 5040
   • D) 288
   Show answer & explanation

   Answer: A) 144

   ARTICLE has 7 letters: vowels A, I, E (3) and consonants R, T, C, L (4). The even positions are 2, 4, 6, exactly 3 places, so vowels can be placed in 3! = 6 ways and consonants in the remaining 4 places in 4! = 24 ways. Total = 6 x 24 = 144. 720 is the answer for 'vowels together' (5! x 3!).

9. Question 9

   If ⁿC₈ = ⁿC₁₂, then ⁿC₁₇ is equal to:

   • A) 20
   • B) 1140
   • C) 6840
   • D) 1330
   Show answer & explanation

   Answer: B) 1140

   ⁿCᵣ = ⁿCₛ with r ≠ s implies n = r + s = 8 + 12 = 20. Then ²⁰C₁₇ = ²⁰C₃ = 20 x 19 x 18 / 6 = 1,140. ²⁰P₃ = 6,840 forgets to divide by 3!.