^{n-1}C_3 + ^{n-1}C_4 ^nC_3,then the value | vedclass

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

Question

(C) Using the Pascal's identity,$^{n-1}C_r + ^{n-1}C_{r-1} = ^nC_r$,we have:$^{n-1}C_3 + ^{n-1}C_4 = ^nC_4$.Given the inequality: $^{n-1}C_3 + ^{n-1}C

Vedclass · Accepted Answer

$> 7$

Vedclass · Answer

(C) Using the Pascal's identity,$^{n-1}C_r + ^{n-1}C_{r-1} = ^nC_r$,we have:$^{n-1}C_3 + ^{n-1}C_4 = ^nC_4$.Given the inequality: $^{n-1}C_3 + ^{n-1}C_4 > ^nC_3$.Substituting the identity: $^nC_4 > ^nC_3$.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we get:$\frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!}$.Dividing both sides by $n!$ and simplifying:$\frac{1}{4 \cdot 3!(n-4)!} > \frac{1}{3!(n-3)(n-4)!}$.$\frac{1}{4} > \frac{1}{n-3}$.Since $n-3 > 0$ (as $n \ge 4$ for combinations to be defined),we have:$n-3 > 4 \Rightarrow n > 7$.

$^{n-1}C_3 + ^{n-1}C_4 > ^nC_3$,then the value of $n$ is

Similar Questions

How many numbers can be formed using the digits $0, 1, 2, 3, 4, 5$ that are greater than $3000$ (repetition is not allowed)?

$A$ question paper is divided into two parts $A$ and $B$,and each part contains $5$ questions. The number of ways in which a candidate can answer $6$ questions by selecting at least two questions from each part is:

How many words (with or without meaning) of $3$ distinct English alphabets can be formed?

$\binom{n}{n-r} + \binom{n}{r+1}$,whenever $0 \le r \le n-1$,is equal to

In how many ways can $10$ balls be divided between two boys,if one receives two balls and the other receives eight balls?

Vedclass Test Series

Exam Paper Generator

Online Exam Module