# Solve the following

Question:

Evaluate ${ }^{20} C_{5}+\sum_{r=2}^{5}{ }^{25-r} C_{4}$

Solution:

Given:

${ }^{20} C_{5}+\sum_{r=2}^{5} 25-{ }^{r} C_{4}$

${ }^{20} C_{5}+\sum_{r=2}^{5} 25-{ }^{r} C_{4}$

$={ }^{20} C_{5}+{ }^{23} C_{4}+{ }^{22} C_{4}+{ }^{21} C_{4}+{ }^{20} C_{4}$

$=\left({ }^{20} C_{4}+{ }^{20} C_{5}\right)+{ }^{21} C_{4}+{ }^{22} C_{4}+{ }^{23} C_{4}$

$={ }^{21} C_{5}+{ }^{21} C_{4}+{ }^{22} C_{4}+{ }^{23} C_{4} \quad\left[\because{ }^{n} C_{r-1}+{ }^{n} C_{r}={ }^{n+1} C_{r}\right]$

$=\left({ }^{21} C_{4}+{ }^{21} C_{5}\right)+{ }^{22} C_{4}+{ }^{23} C_{4}$

$={ }^{22} C_{5}+{ }^{22} C_{4}+{ }^{23} C_{4} \quad\left[\because{ }^{n} C_{r-1}+{ }^{n} C_{r}={ }^{n+1} C_{r}\right]$

$={ }^{23} C_{5}+{ }^{22} C_{4}+{ }^{23} C_{4}$

$={ }^{23} C_{5}+{ }^{23} C_{4}$             $\left[\because{ }^{n} C_{r-1}+{ }^{n} C_{r}={ }^{n+1} C_{r}\right]$

$={ }^{24} C_{5}$

$=\frac{25 !}{19 ! 5 !}=\frac{24 \times 23 \times 22 \times 21 \times 20}{5 \times 4 \times 3 \times 2 \times 1}=42504$