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

$^nC_r + ^nC_{r-1}$ is equal to

The number of solutions of the equation $x+y+z=10$, where $x, y$, and $z$ are positive integers, is:

The least value of the natural number $n$ satisfying $C(n, 5)+C(n, 6)>C(n+1, 5)$ is

There are two urns. Urn $A$ has $3$ distinct red balls and urn $B$ has $9$ distinct blue balls. From each urn,two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is

The number of ways five alphabets can be chosen from the alphabets of the word $MATHEMATICS$,where the chosen alphabets are not necessarily distinct,is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module