BG wrote:

Looking at previous posts seems that my logic is wrong but still i will give it a try.

2 VP's are included so we need 4 more people. But another restriction is that SI and P can not be taken together. Then the other 4 members will have to be selected out of 9, which can be done in 126 ways=9C4. But this result should be multiplied by 2 since one time SI is excluded, other P is excluded from the group of 10

So the final outcome comes to be 252

I believe 2 vice presidents is always a must, therefore in one case it is 1) VP VP I x x x (but cannot be P) in other case 2) VP VP P x x x (but cannot be I).

All in all we have 12 persons.

In the first case we have 9 people left to be selected to 3 places. However, we cannot choose President in that place. Therefore, 8 persons are left to be selected in 3 places.

Similarly is the case for the second case, where 9 people are left to be selected in 3 places, however the inspector cannot be selected. We are left with 8 persons to 3 places.

Now 8c4 is the situation when in this group of 6 members we do not have President nor the inspector. Therefore, VP VP x x x x which initially says that there are 4 places and 10 people to be selected, when we throw out

president and inspector we are left with 8 people to be selected to 4 places. Here we have a formula 8c4.

So 8c3+8c3+8c4 is the answer.......