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(b) For $ m$ $ = 5,\,\,\,\,\sum\limits_{i = 0}^5 {\left( \begin{array}{l}\,10\,\\,\,i\end{array} ight)\,\left( \begin{array}{l}\,\,20\,\5 - i\,\en

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$15$

Vedclass · Answer

(b) For $ m$ $ = 5,\,\,\,\,\sum\limits_{i = 0}^5 {\left( \begin{array}{l}\,10\,\\,\,i\end{array} ight)\,\left( \begin{array}{l}\,\,20\,\5 - i\,\end{array} ight)} $

$ = \left( \begin{array}{l}\,10\\,\,0\end{array} ight)\,\,\left( \begin{array}{l}20\\,5\end{array} ight)\,\, + \,\left( \begin{array}{l}10\\,1\end{array} ight)\,\left( \begin{array}{l}20\\,4\end{array} ight)\, + ... + \left( \begin{array}{l}10\\,5\end{array} ight)\,\left( \begin{array}{l}20\\,0\end{array} ight),$

for $m = 10$, $\sum\limits_{i = 0}^{10} {\left( \begin{array}{l}\,10\,\\,\,i\end{array} ight)\,\left( \begin{array}{l}\,\,20\,\10 - i\,\end{array} ight)} $
$ = \left( \begin{array}{l}10\\,0\end{array} ight)\,\left( \begin{array}{l}20\10\end{array} ight) + \left( \begin{array}{l}10\\,1\end{array} ight)\,\left( \begin{array}{l}20\\,9\,\end{array} ight) + \left( \begin{array}{l}10\\,2\end{array} ight)\,\left( \begin{array}{l}20\\,8\,\end{array} ight)$$ + ... + \left( \begin{array}{l}10\,\10\end{array} ight)\,\left( \begin{array}{l}20\\,0\end{array} ight)$,
for m = 15, $\sum\limits_{i = 0}^{15} {\left( \begin{array}{l}\,10\,\\,\,i\end{array} ight)\,\left( \begin{array}{l}\,\,\,20\,\15 - i\,\end{array} ight)} $
$ = \left( \begin{array}{l}\,10\\,\,0\end{array} ight)\,\,\left( \begin{array}{l}20\\,15\end{array} ight)\,\, + \,\left( \begin{array}{l}10\\,1\end{array} ight)\,\left( \begin{array}{l}20\1\,4\end{array} ight)\, + \left( \begin{array}{l}10\\,2\end{array} ight)\,\left( \begin{array}{l}20\\,13\end{array} ight) + .. + \left( {\begin{array}{*{20}{c}}{10}\{10}\end{array}} ight)\,\left( {\begin{array}{*{20}{c}}{20}\5\end{array}} ight)$

and for $m = 20$, $\sum\limits_{i = 0}^{20} {\left( \begin{array}{l}\,10\,\\,\,i\end{array} ight)\,\left( \begin{array}{l}\,\,\,20\,\20 - i\,\end{array} ight)} $
$ = \left( \begin{array}{l}\,10\\,\,0\end{array} ight)\,\,\left( \begin{array}{l}20\20\end{array} ight)\,\, + \,\left( \begin{array}{l}10\\,1\end{array} ight)\,\left( \begin{array}{l}20\1\,9\end{array} ight)\, + ... + \left( \begin{array}{l}10\10\end{array} ight)\,\left( \begin{array}{l}20\\,10\end{array} ight)$

Clearly, the sum is maximum for $m = 15$.

Note that $^{10}{C_r}$ is maximum for $r = 5$ and $^{20}{C_r}$ is maximum for $r = 10$. Note that the single term $^{10}{C_5}\, 	imes {\,^{20}}{C_{10}}\,$(in case m = 15) is greater than the sum

$^{10}{C_0}{\,^{20}}{C_{10}}\, + {\,^{10}}{C_1}{\,^{20}}{C_9}\, + {\,^{10}}{C_2}{\,^{20}}{C_8}\, + .....$

$^{10}{C_8}{\,^{20}}{C_2}\, + {\,^{10}}{C_9}{\,^{20}}{C_1}\, + {\,^{10}}{C_{10}}{\,^{20}}{C_0}$ (in case $m = 10$).

Also the sum in case $m = 10$ is same as that in case $m = 20.$

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