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STD 11 - 6. permutation and combination — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 6. permutation and combination1 Mark
MCQ
The sum $\sum\limits_{i = 0}^m {\left( {\begin{array}{*{20}{c}}{10}\\i\end{array}} \right)} \,\left( {\begin{array}{*{20}{c}}{20}\\{m - i}\end{array}} \right)\,,$ $\left( {{\rm{where}}\,\left( {\begin{array}{*{20}{c}}p\\q\end{array}} \right)\, = 0\,{\rm{if}}\,p < q} \right)$, is maximum when m is
  • A
    $5$
  • ✓
    $15$
  • C
    $10$
  • D
    $20$
✓

Answer

Correct option: B.
$15$
b
(b) For $ m$ $ = 5,\,\,\,\,\sum\limits_{i = 0}^5 {\left( \begin{array}{l}\,10\,\\\,\,i\end{array} \right)\,\left( \begin{array}{l}\,\,20\,\\5 - i\,\end{array} \right)} $

$ = \left( \begin{array}{l}\,10\\\,\,0\end{array} \right)\,\,\left( \begin{array}{l}20\\\,5\end{array} \right)\,\, + \,\left( \begin{array}{l}10\\\,1\end{array} \right)\,\left( \begin{array}{l}20\\\,4\end{array} \right)\, + ... + \left( \begin{array}{l}10\\\,5\end{array} \right)\,\left( \begin{array}{l}20\\\,0\end{array} \right),$

for $m = 10$, $\sum\limits_{i = 0}^{10} {\left( \begin{array}{l}\,10\,\\\,\,i\end{array} \right)\,\left( \begin{array}{l}\,\,20\,\\10 - i\,\end{array} \right)} $
$ = \left( \begin{array}{l}10\\\,0\end{array} \right)\,\left( \begin{array}{l}20\\10\end{array} \right) + \left( \begin{array}{l}10\\\,1\end{array} \right)\,\left( \begin{array}{l}20\\\,9\,\end{array} \right) + \left( \begin{array}{l}10\\\,2\end{array} \right)\,\left( \begin{array}{l}20\\\,8\,\end{array} \right)$$ + ... + \left( \begin{array}{l}10\,\\10\end{array} \right)\,\left( \begin{array}{l}20\\\,0\end{array} \right)$,
for m = 15, $\sum\limits_{i = 0}^{15} {\left( \begin{array}{l}\,10\,\\\,\,i\end{array} \right)\,\left( \begin{array}{l}\,\,\,20\,\\15 - i\,\end{array} \right)} $
$ = \left( \begin{array}{l}\,10\\\,\,0\end{array} \right)\,\,\left( \begin{array}{l}20\\\,15\end{array} \right)\,\, + \,\left( \begin{array}{l}10\\\,1\end{array} \right)\,\left( \begin{array}{l}20\\1\,4\end{array} \right)\, + \left( \begin{array}{l}10\\\,2\end{array} \right)\,\left( \begin{array}{l}20\\\,13\end{array} \right) + .. + \left( {\begin{array}{*{20}{c}}{10}\\{10}\end{array}} \right)\,\left( {\begin{array}{*{20}{c}}{20}\\5\end{array}} \right)$

and for $m = 20$, $\sum\limits_{i = 0}^{20} {\left( \begin{array}{l}\,10\,\\\,\,i\end{array} \right)\,\left( \begin{array}{l}\,\,\,20\,\\20 - i\,\end{array} \right)} $
$ = \left( \begin{array}{l}\,10\\\,\,0\end{array} \right)\,\,\left( \begin{array}{l}20\\20\end{array} \right)\,\, + \,\left( \begin{array}{l}10\\\,1\end{array} \right)\,\left( \begin{array}{l}20\\1\,9\end{array} \right)\, + ... + \left( \begin{array}{l}10\\10\end{array} \right)\,\left( \begin{array}{l}20\\\,10\end{array} \right)$

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 $^{10}{C_5}\, \times {\,^{20}}{C_{10}}\,$(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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