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?
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.