Evaluate ${ }^{20} C_{5}+\sum_{r=2}^{5}{ }^{25-r} C_{4}$
Given:
${ }^{20} C_{5}+\sum_{r=2}^{5} 25-{ }^{r} C_{4}$
${ }^{20} C_{5}+\sum_{r=2}^{5} 25-{ }^{r} C_{4}$
$={ }^{20} C_{5}+{ }^{23} C_{4}+{ }^{22} C_{4}+{ }^{21} C_{4}+{ }^{20} C_{4}$
$=\left({ }^{20} C_{4}+{ }^{20} C_{5}\right)+{ }^{21} C_{4}+{ }^{22} C_{4}+{ }^{23} C_{4}$
$={ }^{21} C_{5}+{ }^{21} C_{4}+{ }^{22} C_{4}+{ }^{23} C_{4} \quad\left[\because{ }^{n} C_{r-1}+{ }^{n} C_{r}={ }^{n+1} C_{r}\right]$
$=\left({ }^{21} C_{4}+{ }^{21} C_{5}\right)+{ }^{22} C_{4}+{ }^{23} C_{4}$
$={ }^{22} C_{5}+{ }^{22} C_{4}+{ }^{23} C_{4} \quad\left[\because{ }^{n} C_{r-1}+{ }^{n} C_{r}={ }^{n+1} C_{r}\right]$
$={ }^{23} C_{5}+{ }^{22} C_{4}+{ }^{23} C_{4}$
$={ }^{23} C_{5}+{ }^{23} C_{4}$ $\left[\because{ }^{n} C_{r-1}+{ }^{n} C_{r}={ }^{n+1} C_{r}\right]$
$={ }^{24} C_{5}$
$=\frac{25 !}{19 ! 5 !}=\frac{24 \times 23 \times 22 \times 21 \times 20}{5 \times 4 \times 3 \times 2 \times 1}=42504$
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