Question

how many different committees can be formed from 11 teachers and 50 students if the committee consists of 3 teachers and 4 students?

Answer 1

37999500 committees can be formed from 11 teachers and 50 students.

Solution:

Using combination formula, $\bold{n C_{r}=\frac{n !}{(n-r) ! r !}}$

‘n’ represents total number of items and ‘r’ indicates number of item being chosen.

$\begin{array}{l}{\text { (3 out of } 11 \text { teachers } \times 4 \text { out of } 50 \text { students })} \\ \\{(\mathrm{n}=11, \mathrm{r}=3) \times(\mathrm{n}=50, \mathrm{r}=4)} \\\\ {11 C_{3} \times 50 C_{4}=\left(\frac{11 !}{3 !(11-3) !}\right) \times\left(\frac{50 !}{4 !(50-4) !}\right)} \\\\ {=\left(\frac{11 !}{3 ! 8 !}\right) \times\left(\frac{50 !}{4 ! 46 !}\right)=165 \times 230300=37999500}\end{array}$