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Hi maam, can u pls help me with this problem.

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.

Suppose there are n boys, and you need to seat r of them. There is no repetition (i.e., a boy cannot occupy more than one seat), but the order of seating matters. The number of arrangements is n!/(n-r)!.

In this case, n=6 and r=6.

6!/(6-6)! = 6!/0! = 6!/1 = 720

There are 720 ways to seat the boys in the odd chairs.

Similarly, there are 720 ways to seat the girls in the even chairs.

This gives a total of 720×720 = 518,400 arrangements in which boys occupy only the odd chairs.

More information:

http://www.mathsisfun.com/combinatorics/combinations-permutations.html

Jump down to case #2, permutations without repetition.

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