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

The number of $5$ card combinations out of a deck of $52$ cards such that there is exactly one ace in each combination is:

The number of ways to select one or more books from $6$ books is:

If ${ }^{15} C_4+{ }^{15} C_5+{ }^{16} C_6+{ }^{17} C_7+{ }^{18} C_8={ }^{19} C_{r}$,then the value of $r$ is equal to

Two players $A$ and $B$ play a series of games of badminton. The player, who wins $5$ games first, wins the series. Assuming that no game ends in a draw, the number of ways, in which player $A$ wins the series is . . . . . . .

In a conference of $8$ persons,if each person shakes hands with every other person exactly once,then the total number of handshakes is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module