This is a question relatedc to permutations and combinations.

Dear Student,

 

The correct answer is option B.

 

Statement 1: We separate the youngest and the eldest boys and divide the remaining boys into groups of 2, 3, 3.

This can be done in /$(\frac{\fact{8}}{\fact{2}\fact{2}\fact{3}\fact{3}})/$ ways. Now choose one of the groups having 3 members in 2 ways and put the eldest person in that group. The youngest boy goes in the group with 2 members. Thus the final answer turns out to be /$(\frac{\fact{8}}{\fact{2}\fact{3}\fact{3}})/$.

 

Statement 2: Permute the (2m+n) objects and divide by m!, m! and n! as they are to be grouped. But this results in double counting as we can count either of the groups containing m elemnts first, thus we must divide by 2. So the answer is /$(\frac{\fact{2m+n}}{2\fact{m}^2\fact{n}})/$

 

 

Therefore both the statments are true but Statement 2 is not the correct explanation of Statement 1.

 

Regards, 

Vedant