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# Word Problems/Permutation problem

Question
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.

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.

http://www.mathsisfun.com/combinatorics/combinations-permutations.html
Jump down to case #2, permutations without repetition.

Word Problems

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#### Janet Yang

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Word problems are my favorite type of math questions! I would not feel comfortable answering questions that require specialized knowledge (Physics, Statistics, etc.) because I have not studied these in depth.

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