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

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(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-

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$15$ or $8$

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(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 $n$.

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