Find the value of _ r=1 ^4 ^ 21-r C_4+ ^ 17 C_5 - Vidyadip

Question

Find the value of $\sum_{r=1}^4{ }^{21-r} C_4+{ }^{17} C_5$

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Answer

$\sum_{\mathrm{r}=1}^4{ }^{(21-\mathrm{r})} \mathrm{C}_4+{ }^{17} \mathrm{C}_5=\left({ }^{20} \mathrm{C}_4+{ }^{19} \mathrm{C}_4+{ }^{18} \mathrm{C}_4+{ }^{17} \mathrm{C}_4\right)+{ }^{17} \mathrm{C}_5$
$=\left({ }^{20} \mathrm{C}_4+{ }^{19} \mathrm{C}_4+{ }^{18} \mathrm{C}_4+{ }^{18} \mathrm{C}_5-{ }^{17} \mathrm{C}_5\right)+{ }^{17} \mathrm{C}_5\text { ... }\left[{ }^n \mathrm{C}_{\mathrm{r}}+{ }^n \mathrm{C}_{\mathrm{r}-1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}\right]$
$={ }^{20} \mathrm{C}_4+{ }^{19} \mathrm{C}_4+\left({ }^{18} \mathrm{C}_4+{ }^{18} \mathrm{C}_5\right)$
$={ }^{20} \mathrm{C}_4+{ }^{19} \mathrm{C}_4+{ }^{19} \mathrm{C}_5$
$={ }^{20} \mathrm{C}_4+{ }^{20} \mathrm{C}_5$
$={ }^{21} \mathrm{C}_5$
$\sum_{\mathrm{r}=1}^4{ }^{(21-\mathrm{r})} \mathrm{C}_4+{ }^{17} \mathrm{C}_5={ }^{21} \mathrm{C}_5$

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