Question

A team of four 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. $220$
• B. $182$
• C. $226$
• D. $495$

Answer 1

Answer: (A), (C)

The correct options are
A $220$
C $226$

Let $S_1$ and $S_2$ refuse to be together and $S_3$ and $S_4$ want to be together only. The total number of ways when $S_3$ and $S_4$ are selected is ${(}^8{C}_2{+}^2{C}_1{\times }^8{C}_1)=44$. The total ways when $S_3$ and $S_4$ are not selected is ${(}^8{C}_4{+}^2{C}_1{\times }^8{C}_3)=182$. Thus, the total number of ways is $44+182=226$.