# The value of

Question:

The value of $\sum_{r=0}^{20}{ }^{50-r} C_{6}$ is equal to :

1. ${ }^{51} \mathrm{C}_{7}+{ }^{30} \mathrm{C}_{7}$

2. ${ }^{51} C_{7}-{ }^{30} C_{7}$

3. ${ }^{50} \mathrm{C}_{7}-{ }^{30} \mathrm{C}_{7}$

4. ${ }^{50} \mathrm{C}_{6}-{ }^{30} \mathrm{C}_{6}$

Correct Option: , 2

Solution:

$\sum_{\mathrm{r}=0}^{20}{ }^{50-\mathrm{r}} \mathrm{C}_{6}={ }^{50} \mathrm{C}_{6}+{ }^{49} \mathrm{C}_{6}+{ }^{48} \mathrm{C}_{6}+\ldots . .+{ }^{30} \mathrm{C}_{6}$

$={ }^{50} \mathrm{C}_{6}+{ }^{49} \mathrm{C}_{6}+\ldots . .+{ }^{31} \mathrm{C}_{6}+\left({ }^{30} \mathrm{C}_{6}+{ }^{30} \mathrm{C}_{7}\right)-{ }^{30} \mathrm{C}_{7}$

$={ }^{50} \mathrm{C}_{6}+{ }^{49} \mathrm{C}_{6}+\ldots . .+\left({ }^{31} \mathrm{C}_{6}+{ }^{31} \mathrm{C}_{7}\right)-{ }^{30} \mathrm{C}_{7}$

$={ }^{50} \mathrm{C}_{6}+{ }^{50} \mathrm{C}_{7}-{ }^{30} \mathrm{C}_{7}$

$={ }^{51} \mathrm{C}_{7}-{ }^{30} \mathrm{C}_{7}$

${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}$