 # A committee of 5 is to be formed out of 6 gents and 4 ladies.

Question:

A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many ways can this be done, when

(i) at least 2 ladies are included?

(ii) at most 2 ladies are included?

Solution:

Since the committee of 5 is to be formed from 6 gents and 4 ladies.

(i) Forming a committee with at least 2 ladies

Here the possibilities are

(i) 2 ladies and 3 gents

(ii) 3 ladies and 2 gents

(iii) 4 ladies and 1 gent

The number of ways they can be selected

$={ }^{4} \mathrm{C}_{2}$ $\times$ ${ }^{6} \mathrm{C}_{3}+{ }^{4} \mathrm{C}_{3}$ $\times$ ${ }^{6} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{4}$ $\times$ ${ }^{6} \mathrm{C}_{1}$

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

$=186$ ways

(ii) The number of ways in this case is

1. 0 ladies and 5 gents

2. 1 lady and 4 gents

3. 2 ladies and 3 gents.

The total ways are

$={ }^{4} \mathrm{C}_{0} \times{ }^{6} \mathrm{C}_{5}+{ }^{4} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{2} \times{ }^{6} \mathrm{C}_{3}$

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

$=186$ ways.