Match List-I with List-II
List-I (Expressions):
(A) Find x
(B) Evaluate: , n = 6, r = 2
(C) If , fi

Q: What is the correct answer for CUET UG Quantitative Aptitude?

The correct answer is C. (A) $1/6! + 1/7! = x/8!$ Multiply by $8!$: $8!/6! + 8!/7! = x$ $8 \times 7 + 8 = x$ $56 + 8 = x \Rightarrow x = 64$. So, (A) matches with (III). (B) Evaluate: $\frac{n!}{(n-r)!}$ for n = 6, r = 2. This is $^nP_r$. $^6P_2 = 6 \times 5 = 30$. So, (B) matches with (IV). (C) If $^nC_9 = ^nC_8$, then $n = 9 + 8 = 17$. We need to find $^nC_{17}$, which is $^17C_{17} = 1$. So, (C) matches with (I). (D) $

Q: Match List-I with List-II List-I (Expressions): (A) Find x (B) Evaluate: , n = 6, r = 2 (C) If , find . (D) List-II (Values): (I) 1 (II) 100 (III) 64 (IV) 30 Choose the correct answer from the…

The correct answer is C. (A) $1/6! + 1/7! = x/8!$ Multiply by $8!$: $8!/6! + 8!/7! = x$ $8 \times 7 + 8 = x$ $56 + 8 = x \Rightarrow x = 64$. So, (A) matches with (III). (B) Evaluate: $\frac{n!}{(n-r)!}$ for n = 6, r = 2. This is $^nP_r$. $^6P_2 = 6 \times 5 = 30$. So, (B) matches with (IV). (C) If $^nC_9 = ^nC_8$, then $n = 9 + 8 = 17$. We need to find $^nC_{17}$, which is $^17C_{17} = 1$. So, (C) matches with (I). (D) $

Q: What are the options for this CUET UG Quantitative Aptitude question?

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

Question

Match List-I with List-II
List-I (Expressions):
(A)  Find x
(B) Evaluate: , n = 6, r = 2
(C) If , find .
(D) 
List-II (Values):
(I) 1
(II) 100
(III) 64
(IV) 30
Choose the correct answer from the options given below:

FirstInTest · Accepted Answer

(A) $1/6! + 1/7! = x/8!$
Multiply by $8!$: $8!/6! + 8!/7! = x$
$8 	imes 7 + 8 = x$
$56 + 8 = x \Rightarrow x = 64$. So, (A) matches with (III).
(B) Evaluate: $\frac{n!}{(n-r)!}$ for n = 6, r = 2. This is $^nP_r$.
$^6P_2 = 6 	imes 5 = 30$. So, (B) matches with (IV).
(C) If $^nC_9 = ^nC_8$, then $n = 9 + 8 = 17$. We need to find $^nC_{17}$, which is $^17C_{17} = 1$. So, (C) matches with (I).
(D) $^6P_3 - ^5P_2$
$^6P_3 = 6 	imes 5 	imes 4 = 120$
$^5P_2 = 5 	imes 4 = 20$
$120 - 20 = 100$. So, (D) matches with (II).
Final Matching: (A) - (III), (B) - (IV), (C) - (I), (D) - (II). This corresponds to option 3.

Answer

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

Answer

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

Answer

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

Answer

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

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