Question

How many committees of 5 members each can be formed from 8 official and 4 non-official members in the following cases:
Each consisting of 3 official and 2 non-official members.
(b)Each contains at least two non-official members.
(c)A particular official member is never included.
(d)A particular non-official member is always included.

Answer 1

In this question we must bear in mind that we have only to form committees. We are not concerned with the arrangement of officials or non-officials.
8 officials, 4 non-officials.
(a) 3 officials and 2 non-officials.
3 officials out of 8 can be selected by (s) (iii), in
${}^8{C}_3$ = $\frac{8!}{5!3!}=\frac{8\times 7\times 6}{1\times 2\times 3}$ = 56 ways
2 non-officials out of 4 can be selected in
${}^4{C}_2=\frac{4.3}{1.2}$ = 6 ways
By fundamental theorem the number of ways in which the committee can be formed is 56 x 6=336
(b) At least two non-officials = Total - One non-official - No non-off
${}^{12}{C}_5$ - (${}^8{C}_4{.}^4{C}_1$) - ${}^8{C}_5$
= 792 - 280 - 56 = 456
(c) A particular official never included.
Required no. of ways
= ${}^{12-1}{C}_5$ = ${}^{1}1C_5$ = $\frac{\mathrm{11.10.9.8.7}}{\mathrm{1.2.3.4.5}}$ = 462
[Here p = 1, r = 5 and n = 12]
(d) A particular non-ofificial is to be always included
The required number of ways
= ${}^{11}{C}_4$ = $\frac{\mathrm{11.10.9.8}}{\mathrm{1.2.3.4}}$ = 330
[p = 1,r = 5 and n = 12]