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 ⁴C₁ ways

Therefore, 5 will be chosen form 8 clerks in ⁸C₅ ways.

Thus by multiplication principle , we get

Total no. of ways in 1 case is ⁴C₁ ⁸C_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 ¹²C₆ ways.

And cases that would have only clerks would be i.e. selecting 6 from 8 clerks in ⁸C₆ ways.

¹²C₆ - ⁸C₆ ways.

Applying ⁿC_r =

⇒924 - 28 ways

= 896 ways