Target Exam

CUET

Subject

General Aptitude Test

Chapter

Quantitative Reasoning

Topic

Permutation & Combination

Question:

Match List-I with List-II

List-I

List-II

(A) ${^{75}P}_2-{^{75}C}_2$

(I) 504

(B) ${^5P}_5-{^{10}C}_3$

(II) 6

(C) ${^{16}C}_{13}-{^8C}_3$

(III) 2775

(D) ${^nP}_4=360$, then find $n$ 

(IV) 0

Choose the correct answer from the options given below:

Options:

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Correct Answer:

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Explanation:

The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

List-I

List-II

(A) ${^{75}P}_2-{^{75}C}_2$

(III) 2775

(B) ${^5P}_5-{^{10}C}_3$

(IV) 0

(C) ${^{16}C}_{13}-{^8C}_3$

(I) 504

(D) ${^nP}_4=360$, then find $n$ 

(II) 6

(A) $^{75}P_2 - ^{75}C_2$

  • $^{75}P_2 = 75 \times 74 = 5550$
  • $^{75}C_2 = \frac{75 \times 74}{2 \times 1} = 2775$
  • Difference: $5550 - 2775 = \mathbf{2775}$
  • Matches: (III)

(B) $^5P_5 - ^{10}C_3$

  • $^5P_5 = 5! = 120$
  • $^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$
  • Difference: $120 - 120 = \mathbf{0}$
  • Matches: (IV)

(C) $^{16}C_{13} - ^8C_3$

  • $^{16}C_{13} = ^{16}C_{16-13} = ^{16}C_3 = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$
  • $^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
  • Difference: $560 - 56 = \mathbf{504}$
  • Matches: (I)

(D) $^nP_4 = 360$

  • We need to find $n$ such that $n(n-1)(n-2)(n-3) = 360$.
  • Testing $n = 6$: $6 \times 5 \times 4 \times 3 = 360$.
  • Matches: (II)