**Question:**

Out of 6 teachers and 8 students, a committee of 11 is being formed. In how many ways can this be done, if the committee contains

(i) exactly 4 teachers?

(ii) at least 4 teachers?

**Solution:**

Since the committee of 11 is to be formed from 6 teachers and 8 students.

(i) Forming a committee with exactly 4 teachers

Choosing 4 teachers out of 6 in ${ }^{6} \mathrm{C}_{4}$ ways.

Remaining 7 from 8 students in ${ }^{8} \mathrm{C} 7$ ways.

Thus, total ways in (i) are ${ }^{6} \mathrm{C}_{4} \times{ }^{8} \mathrm{C}_{7}$ ways.

(ii) The number of ways in this case is

1. 4 teachers and 7 students

2. 5 teachers and 6 students

3. 6 teachers and 5 students

$={ }^{6} \mathrm{C}_{4} \times{ }^{8} \mathrm{C}_{7}+{ }^{6} \mathrm{C}_{5} \times{ }^{8} \mathrm{C}_{6}+{ }^{6} \mathrm{C}_{6} \times{ }^{8} \mathrm{C}_{5}$

Applying ${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$

$=344$ ways