**Question:**

From 4 officers and 8 clerks, in how many ways can 6 be chosen

(i) to include exactly one officer,

(ii) to include at least one officer?

**Solution:**

The team of 6 has to be chosen from 4 officers and 8 clerks. There are some restrictions which are

1. To include exactly one officer

In this case,

One officer will be chosen from 4 in ${ }^{4} \mathrm{C}_{1}$ ways

Therefore, 5 will be chosen form 8 clerks in ${ }^{8} \mathrm{C}_{5}$ ways.

Thus by multiplication principle, we get

Total no. of ways in 1 case is ${ }^{4} C_{1} \times{ }^{8} C_{5}$

2. To include at least one officer

In this case, there will be subcases for selection which is as follows.

(i) One officer and 5 clerks

(ii) Two officers and 4 clerks

(iii) Three officers and 3 clerks

(iv) Four officers and 2 clerks

Or

The required case of at least on officer would be

= Total cases – cases having only clerks

Now,

The total case would be choosing 6 out of 12 in ${ }^{12} \mathrm{C} 6$ ways.

And cases that would have only clerks would be i.e. selecting 6 from 8 clerks in ${ }^{8} \mathrm{C}_{6}$ ways.

$\Rightarrow{ }^{12} C_{6}-{ }^{8} C_{6}$ ways.

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

$\Rightarrow 924-28$ ways

$=896$ ways