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6.Permutation and Combination
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$\left( {\begin{array}{*{20}{c}}n\\{n - r}\end{array}} \right)\, + \,\left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right)$, whenever $0 \le r \le n - 1$is equal to
A
$\left( {\begin{array}{*{20}{c}}n\\{r - 1}\end{array}} \right)$
B
$\left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right)$
C
$\left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right)$
D
$\left( {\begin{array}{*{20}{c}}{n + 1}\\{r + 1}\end{array}} \right)$
Solution
(d)$\left( \begin{array}{l}\,\,\,n\,\\\,n – r\end{array} \right)$+$\left( \begin{array}{l}\,\,\,n\,\\r + 1\end{array} \right)$ = $^n{C_{n – r}}{ + ^n}{C_{r + 1}}$
$ \Rightarrow {\,^n}{C_r}\, + {\,^n}{C_{r + 1}}$ = $^{n + 1}{C_{r + 1}} = \left( {\begin{array}{*{20}{c}}{n + 1}\\{r + 1}\end{array}} \right)$.
Standard 11
Mathematics
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