#### In a lottery 100 females and 50 males participated. In how many ways can the rejection list of 50 members be made regardless of female and male participants such that 20 participants are marked to be rejected?Option: 1 $\frac{50!}{30!100!}$Option: 2 $\frac{130!}{50!100!}$Option: 3 $\frac{150!}{30!100!}$Option: 4 $\frac{130!}{30!100!}$

Note the following:

• The formula for the combination for the selection of the $\mathrm{x}$ items from the $\mathrm{y}$different items is $\mathrm{=^{y}C_{x}=\frac{y!}{x!\left ( y-x \right )!}}$

• The restricted combination for the selection of the $\mathrm{r}$  items from the $\mathrm{n}$different items with $\mathrm{k}$ particular things always included is $\mathrm{=^{n-k}C_{r-k}}$

Since 20 participants are marked to be rejected, the following is evident.

• The number from which the restricted combination is to be made is .

$\mathrm{=n-k=\left ( 100+50 \right )-20=130}$

• The number with which the restricted combination is to be made is

$\mathrm{=r-k=50-20=30}$

Therefore, the required restricted combination is

$\mathrm{=^{n-k}C_{r-k}}$

$\mathrm{=^{130}C_{30}}$

$\mathrm{=\frac{130!}{30!100!}}$