If binom{n-1}{4},binom{n-1}{5},and binom{n-1} | vedclass

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

Question

(A) Since $\binom{n-1}{4}$,$\binom{n-1}{5}$,and $\binom{n-1}{6}$ are in arithmetic progression,we have: $2 \binom{n-1}{5} = \binom{n-1}{4} + \binom{n-

Vedclass · Accepted Answer

$15$ or $8$

Vedclass · Answer

(A) Since $\binom{n-1}{4}$,$\binom{n-1}{5}$,and $\binom{n-1}{6}$ are in arithmetic progression,we have: $2 \binom{n-1}{5} = \binom{n-1}{4} + \binom{n-1}{6}$ Using the property $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$,we can rewrite the equation: $\binom{n-1}{4} + \binom{n-1}{5} + \binom{n-1}{5} + \binom{n-1}{6} = 4 \binom{n-1}{5}$ $\binom{n}{5} + \binom{n}{6} = 4 \binom{n-1}{5}$ $\binom{n+1}{6} = 4 \binom{n-1}{5}$ Expanding the combinations: $\frac{(n+1)n(n-1)(n-2)(n-3)(n-4)}{6!} = 4 \frac{(n-1)(n-2)(n-3)(n-4)(n-5)}{5!}$ $\frac{(n+1)n}{6} = 4(n-5)$ $n^2 + n = 24(n-5)$ $n^2 - 23n + 120 = 0$ $(n-15)(n-8) = 0$ Thus,$n = 15$ or $n = 8$.

If $\binom{n-1}{4}$,$\binom{n-1}{5}$,and $\binom{n-1}{6}$ are in arithmetic progression,find $n$.

Similar Questions

$A$ committee of $11$ members is to be formed from $8$ males and $5$ females. If $m$ is the number of ways the committee is formed with at least $6$ males and $n$ is the number of ways the committee is formed with at least $3$ females,then:

The greatest positive integer which divides $(n+16)(n+17)(n+18)(n+19)$,for all positive integers $n$,is

$A$ committee of $7$ members is to be formed from $9$ boys and $4$ girls. In how many ways can this be done if the committee consists of at most $3$ girls?

If ${ }^{n-3} C_r + B \cdot { }^{n-3} C_{r-1} + B^{\prime} \cdot { }^{n-3} C_{r-2} + { }^{n-3} C_{r-3} = { }^n C_r$ holds for all $n \geq r \geq 3$,then $(B, B^{\prime}) = $.

If $4$ red balls and $5$ green balls are selected from $n$ balls,and the sum of both selections is greater than ${}^{n+1}C_4$,then the value of $n$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module