If binom{10}{2} + binom{10}{3} + binom{11}{4} | vedclass

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(C) We use the Pascal's identity: $\binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r}$.Given expression: $\binom{10}{2} + \binom{10}{3} + \binom{11}{4} +

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) We use the Pascal's identity: $\binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r}$.Given expression: $\binom{10}{2} + \binom{10}{3} + \binom{11}{4} + \binom{12}{5} + \binom{13}{6} = \binom{14}{r}$.Applying the identity to the first two terms: $\binom{10}{2} + \binom{10}{3} = \binom{11}{3}$.Now the expression becomes: $\binom{11}{3} + \binom{11}{4} + \binom{12}{5} + \binom{13}{6} = \binom{14}{r}$.Applying the identity again: $\binom{11}{3} + \binom{11}{4} = \binom{12}{4}$.Now the expression becomes: $\binom{12}{4} + \binom{12}{5} + \binom{13}{6} = \binom{14}{r}$.Applying the identity again: $\binom{12}{4} + \binom{12}{5} = \binom{13}{5}$.Now the expression becomes: $\binom{13}{5} + \binom{13}{6} = \binom{14}{r}$.Applying the identity one last time: $\binom{13}{5} + \binom{13}{6} = \binom{14}{6}$.Comparing $\binom{14}{6} = \binom{14}{r}$,we get $r = 6$.

If $\binom{10}{2} + \binom{10}{3} + \binom{11}{4} + \binom{12}{5} + \binom{13}{6} = \binom{14}{r}$,then $r = \dots$

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