For integers $n$ and $r$, let $\left(\begin{array}{l}n \ r\end{array} ight)=\left\{\begin{array}{c

Question

For integers $n$ and $r$, let $\left(\begin{array}{l}n \ r\end{array}ight)=\left\{\begin{array}{cc}{ }^{n} C_{r}, & 	ext { if } n \geq r \geq 0 \ 0, & 	ext { otherwise }\end{array}ight.$ The maximum value of $k$ for which the sum $\sum_{i=0}^{k}\left(\begin{array}{c}10 \ i\end{array}ight)\left(\begin{array}{c}15 \ k-i\end{array}ight)+\sum_{i=0}^{k+1}\left(\begin{array}{c}12 \ i\end{array}ight)\left(\begin{array}{c}13 \ k+1-i\end{array}ight)$ exists, is equal to

Professor PC Thomas Classes · Answer

(24) :\ $\sum_{i=0}^{k}\binom{10}{i}\binom{15}{k-i}+\sum_{i=0}^{k+1}\binom{12}{i}\binom{13}{k+1-i}$\ $=\binom{10}{0}\binom{15}{k}+\binom{10}{1}\binom{15}{k-1}+\ldots .+\binom{10}{k}\binom{15}{0}$\ $+\binom{12}{0}\binom{13}{k+1}+\binom{12}{1}\binom{13}{k}+\ldots .+\binom{12}{k+1}\binom{13}{0}$\ $=$ Coeff. of $x^{k}$ in $(1+x)^{10}(1+x)^{15}$Coeff. of $x^{k+1}$ in $(1+x)^{12}(1+x)^{13}$\ $=$ Coeff. of $x^{k}$ in $(1+x)^{25}+$ Coeff. of $x^{k+1}$ in $(1+x)^{25}={ }^{25} C_{k}+{ }^{25} C_{k+1}$\ Now, for sum to be exists $0 \leq k \leq 25$ and $0 \leq k+1 \leq 25 \Rightarrow 0 \leq k \leq 24$\ Thus, maximum value of $k$ is 24 .