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

Q: What is the correct answer for CUET UG Mathematics?

The correct answer is C. (A) $1/6! + 1/7! = x/8!$ Multiply by $8!$: $8!/6! + 8!/7! = x$ $(8 \times 7) + 8 = x$ $56 + 8 = x$ $x = 64$. So (A) matches with (III). (B) Evaluate: $\frac{n!}{(n-r)!}$ for $n=6, r=2$ This is ${}^{6}P_2 = 6 \times 5 = 30$. So (B) matches with (IV). (C) If ${}^{n}C_9 = {}^{n}C_8$, then $n = 9+8 = 17$. We need to find ${}^{n}C_{17} = {}^{17}C_{17} = 1$. So (C) matches with (I). (D) ${}^{6}P_3 -

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 $8!$: $8!/6! + 8!/7! = x$ $(8 \times 7) + 8 = x$ $56 + 8 = x$ $x = 64$. So (A) matches with (III). (B) Evaluate: $\frac{n!}{(n-r)!}$ for $n=6, r=2$ This is ${}^{6}P_2 = 6 \times 5 = 30$. So (B) matches with (IV). (C) If ${}^{n}C_9 = {}^{n}C_8$, then $n = 9+8 = 17$. We need to find ${}^{n}C_{17} = {}^{17}C_{17} = 1$. So (C) matches with (I). (D) ${}^{6}P_3 -

Q: What are the options for this CUET UG Mathematics 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 $8!$: $8!/6! + 8!/7! = x$
$(8 	imes 7) + 8 = x$
$56 + 8 = x$
$x = 64$. So (A) matches with (III).

(B) Evaluate: $\frac{n!}{(n-r)!}$ for $n=6, r=2$
This is ${}^{6}P_2 = 6 	imes 5 = 30$. So (B) matches with (IV).

(C) If ${}^{n}C_9 = {}^{n}C_8$, then $n = 9+8 = 17$. We need to find ${}^{n}C_{17} = {}^{17}C_{17} = 1$. So (C) matches with (I).

(D) ${}^{6}P_3 - {}^{5}P_2$
${}^{6}P_3 = 6 	imes 5 	imes 4 = 120$
${}^{5}P_2 = 5 	imes 4 = 20$
$120 - 20 = 100$. 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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