In how many ways 3 girls and 3 boys be seated in a row of 6 chairs if 1 boy and 1 girl refused to sit next to each other?

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Answer 1 · 2018-06-26T07:05:41

$color(blue)(480 " Ways")$

First we find the number of permutations of the 3 girls and 3 boys filling the 6 chairs, this is not accounting for the two who won't sit together.

$color(white)(0)^6P_6=(6!)/((6-6)!)=6xx5...2xx1=720$

For the number of ways with the two who refuse to sit together being put together:

If the said boy is in seat 1 and girl in seat 2.

There are $4!$ ways for the others to fill the remaining chairs.

If the said boy is in seat 2 and girl in seat 1.

There are $4!$ ways for the others to fill the remaining chairs.

If we do this for seat 2 & 3, 3 & 4 etc. We find we have:

$10 xx 4!$ ways of the said boy and girl being next to each other.

We therefore need to subtract this from $6!$

$6!-4! =480$