If ^nC_3 + ^nC_4 ^{n+1}C_3,then | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We use the Pascal's identity: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.Applying this to the given inequality: $^nC_3 + ^nC_4 = ^{n+1}C_4$.So,the inequality

Vedclass · Accepted Answer

$n > 6$

Vedclass · Answer

(A) We use the Pascal's identity: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.Applying this to the given inequality: $^nC_3 + ^nC_4 = ^{n+1}C_4$.So,the inequality becomes: $^{n+1}C_4 > ^{n+1}C_3$.Expanding the combinations: $\frac{(n+1)!}{4!(n-3)!} > \frac{(n+1)!}{3!(n-2)!}$.Dividing both sides by $(n+1)!$ and simplifying the factorials:$\frac{1}{4 	imes 3!(n-3)!} > \frac{1}{3!(n-2)(n-3)!}$.$\frac{1}{4} > \frac{1}{n-2}$.Since $n > 3$ for the combinations to be defined,$n-2$ is positive,so we can cross-multiply:$n - 2 > 4$.$n > 6$.

If $^nC_3 + ^nC_4 > ^{n+1}C_3$,then

Similar Questions

If $^nC_{12} = ^nC_6$,then $^nC_2 = $

$A$ student is to answer $10$ out of $13$ questions in an examination such that he must choose at least $4$ from the first $5$ questions. The number of choices available to him is

The number of numbers that can be formed with the help of the digits $1, 2, 3, 4, 3, 2, 1$ so that odd digits always occupy odd places,is

The number of ways in which $35$ apples can be distributed among $3$ boys so that each can have any number of apples,is

The number of $4$-digit even numbers that can be formed using $0, 1, 2, 3, 4, 5, 6$ without repetition is

Vedclass Test Series

Exam Paper Generator

Online Exam Module