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Hi sir, can u help me answer this?

6 boys and 6 girls are to be sitted in a row of 11 chairs such that the boys shall occupy only the odd numbers, find the number of all possible ways.

Tnx in adv! Just wanna know how to answer this.

All 6 boys will have seats, that would be 6 choose 6, which is 1.

As far as arranging them, there would be 6! ways to do it.

The number 6! is 1*2*3*4*5*6 = 720.

As far as the 6 girls, only 5 shall be selected, and 6 choose 5 is 6.

As far as the arrangement, there are 5! ways to do it, and 5! = 1*2*3*4*5 = 120.

Now note that when this is multiplied by 6, the answer is 720, which is the same as if the 6th girl had a chair at the end.

When combined, that gives 720 * 720 = 518,400.

Since they could be seated in the same way left to right and right to left,

does the direction at which the chairs are looked at make a difference?

If not, divide the answer 518,400 by 2 and get 259,200.

Consider what would happen if we had 3 people, A, B and C.

If we had a chair for each, the total combinations would be ABC, ACB, BAC, BCA, CAB, and CBA.

That is a total of 6. If we only had two chairs, the combinations would be

AB, AC, BA, BC, CA, and CB, and that would still be 6.

Again, if ABC was considered the same as CBA (ABC in reverse order),

there would only be 3 arrangements for each.

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