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) ${^8P}_3-{^{10}C}_3$

(I) 6

(B) ${^8P}_5$

(II) 21

(C) ${^nP}_4=360$, then find $n$.

(III) 216

(D) ${^nC}_2 = 210$, find $n$.

(IV) 6720

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) ${^8P}_3-{^{10}C}_3$

(III) 216

(B) ${^8P}_5$

(IV) 6720

(C) ${^nP}_4=360$, then find $n$.

(I) 6

(D) ${^nC}_2 = 210$, find $n$.

(II) 21

(A) $^8P_3 - ^{10}C_3 \rightarrow$ (III) 216

  • Calculate $^8P_3$: $8 \times 7 \times 6 = 336$
  • Calculate $^{10}C_3$: $\frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$
  • Subtract: $336 - 120 = \mathbf{216}$

(B) $^8P_5 \rightarrow$ (IV) 6720

  • Calculation: $8 \times 7 \times 6 \times 5 \times 4 = \mathbf{6720}$

(C) $^nP_4 = 360 \rightarrow$ (I) 6

  • The expression for $^nP_4$ is $n(n-1)(n-2)(n-3) = 360$.
  • We look for four consecutive integers whose product is 360.
  • Testing $n = 6$: $6 \times 5 \times 4 \times 3 = 360$.
  • Therefore, $n = 6$.

(D) $^nC_2 = 210 \rightarrow$ (II) 21

  • The expression for $^nC_2$ is $\frac{n(n-1)}{2} = 210$.
  • Multiply by 2: $n(n-1) = 420$.
  • Since $21 \times 20 = 420$, we find that $n = 21$