The least value of a natural number n such th | vedclass

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(C) Given the inequality: $\binom{n-1}{5} + \binom{n-1}{6} < \binom{n}{7}$.Using the Pascal's identity $\binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r}

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(C) Given the inequality: $\binom{n-1}{5} + \binom{n-1}{6} Using the Pascal's identity $\binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r}$,we have $\binom{n-1}{5} + \binom{n-1}{6} = \binom{n}{6}$.Substituting this into the inequality,we get: $\binom{n}{6} Expanding the combinations: $\frac{n!}{6!(n-6)!} Dividing both sides by $n!$ and simplifying the factorials:$\frac{1}{6!(n-6)(n-7)!} $\frac{1}{n-6} Since $n$ is a natural number,$n-6 > 0$,so $n-6 > 7$.$n > 13$.The least natural number $n$ satisfying this condition is $14$.

The least value of a natural number $n$ such that $\binom{n-1}{5} + \binom{n-1}{6} < \binom{n}{7}$ is:

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