Question

A team of $4$ students is to be selected from a total of $12$ students. The total number of ways in which the team can be selected such that two particular students refuse to be together and other two particular students wish to be together only is equal to

• A. $182$
• B. $210$
• C. $226$
• D. $280$

Answer 1

Answer: (C)

Let two particular students $S_1$ and $S_2$ wish to be together and $S_3$ and $S_4$ refuse to be together .
Case :1

Total ways when $S_1,S_2,S_3$ and $S_4$ all are excluded.
$={}^8{C}_4=\frac{8\times 7\times 6\times 5}{1\times 2\times 3\times 4}=70$

Case: 2

Total ways when $S_1,S_2$ are included but $S_3,S_4$ are excluded.
$={}^8{C}_2=\frac{8\times 7}{1\times 2}=28$

Case: 3

Total ways when $S_1,S_2$ are included and any one out of $S_3$ and $S_4$ is included.
$={}^2{C}_1\times {}^8{C}_1=\frac{2}{1}\times \frac{8}{1}=16$

Case: 4

Total ways when $S_1,S_2$ are excluded and any one out of $S_3$ and $S_4$ is included.
$={}^2{C}_1\times {}^8{C}_3=\frac{2}{1}\times \frac{8\times 7\times 6}{1\times 2\times 3}=112$

$\therefore$ Total number of ways $=70+28+16+112=226$.

Hence, Option C is correct.