If binom{18}{15} + 2binom{18}{16} + binom{17} | vedclass

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(C) We use the identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.Given expression: $\binom{18}{15} + \binom{18}{16} + \binom{18}{16} + \binom{

Vedclass · Accepted Answer

$20$

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(C) We use the identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.Given expression: $\binom{18}{15} + \binom{18}{16} + \binom{18}{16} + \binom{17}{16} + 1 = \binom{n}{3}$.Note that $1 = \binom{17}{17}$.So,$\binom{18}{15} + \binom{18}{16} + \binom{18}{16} + \binom{17}{16} + \binom{17}{17} = \binom{n}{3}$.Using the identity $\binom{17}{16} + \binom{17}{17} = \binom{18}{17}$,we get:$\binom{18}{15} + \binom{18}{16} + \binom{18}{16} + \binom{18}{17} = \binom{n}{3}$.Using the identity $\binom{18}{16} + \binom{18}{17} = \binom{19}{17}$,we get:$\binom{18}{15} + \binom{18}{16} + \binom{19}{17} = \binom{n}{3}$.Using the identity $\binom{18}{15} + \binom{18}{16} = \binom{19}{16}$,we get:$\binom{19}{16} + \binom{19}{17} = \binom{n}{3}$.Using the identity $\binom{19}{16} + \binom{19}{17} = \binom{20}{17}$,we get:$\binom{20}{17} = \binom{n}{3}$.Since $\binom{20}{17} = \binom{20}{20-17} = \binom{20}{3}$,we have $\binom{20}{3} = \binom{n}{3}$.Therefore,$n = 20$.

If $\binom{18}{15} + 2\binom{18}{16} + \binom{17}{16} + 1 = \binom{n}{3}$,then find the value of $n$.

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