Question

In how many ways can 12 boys be seated on two benches x, y such that 6 boys sit on each bench and two of them A, B are to sit on bench x and C, D on the bench y?
(a) ${}^{8}{{C}_{4}}6!6!$
(b) ${}^{8}{{C}_{6}}6!6!$
(c) $6!6!$
(d) ${}^{8}{{C}_{4}}6!$

Answer 1

We will first place the two boys on two benches. Then we will choose from the remaining boys to fill in the places remaining on the benches. Then we will count the ways in which 6 boys can be seated on one bench. Then, by multiplying these, we get the total number of ways for the given arrangement.

Complete step-by-step answer:
We are given that two of the 12 boys, A and B, will sit on bench x. Also, two of the remaining 10 boys, C and D, will sit on bench y. The capacity of each bench is given to be 6. After placing the two boys, A and B on bench x and the places of C and D on bench y, the remaining number of places on each bench is 4. Now, we have to choose from the remaining 8 boys, which 4 will occupy one bench and the rest will occupy the second bench. For choosing these 4 boys, we will use combinations. Choosing 4 out of 8 is ${}^{8}{{C}_{4}}$.
Now, on each bench, the 6 boys can seat themselves in $6!$ ways. So, the final number of ways in which 12 boys be seated on two benches x, y such that 6 boys sit on each bench and two of them A, B are to sit on bench x and C, D on the bench y is ${}^{8}{{C}_{4}}6!6!$.
Hence, the correct option is (a).

So, the correct answer is “Option A”.

Note: It is useful to know the concept of combinations, the notations and the formula associated with it. The important point in this is the way we approach the counting. It is important to know when to use counting, combinations and permutations. We should carefully interpret the given information from the question so that we obtain the correct answer.