# In how many ways can 3 boys and 4 girls sit alternately in a row if a particular boy and a girl are to sit together?

I figured out the solution for alternate arrangement of boys and girls without restrictions. If we fix the positions of the four girls, the three boys can only occupy the spaces between them. Now, we can permute the girls and the boys in their fixed positions, so the number of total permutations is $$4!\times 3!=144$$.

With the added restriction that a girl and a boy are to sit together, I realized that there were two cases: the girl could be at an end of the row or she could be in the middle. If she is at an end, there is only one position the boy can take (just next to her), and $$3!\times 2!$$ ways to permute the other 3 girls and 2 boys. If she is in the middle of the row, there are two positions the boy could take (to her right or left) so the total number of permutations now becomes $$2\times 3!\times 2!$$.

The total number of permutations is $$3!\times 2! + 2\times 2!\times 3! + 2\times 2!\times 3! + 2!\times 3! = 72$$.

While solving this problem I got the idea to fix the boy-girl pair and permute the other persons about them, but could not develop it further. Could someone provide an alternative solution, using this or some other idea?

First, seat the other five. You can do this in $$3!\times 2!=12$$ ways. Now there are six places to seat the special pair; and for each position, there is only one way that they can sit to maintain the girl-boy alternation. So the answer is indeed $$6\times 12=72$$.
See that the sitting pattern must be $$GBGBGBG$$. We construct a combination as follow:
1. The individual boy chooses among $$3$$ choice.
2. The individual girl chooses among the $$2$$ adjacent seats next to the one in the first step.
3. The other $$2$$ boys permute $$2!$$ ways.
4. The other $$3$$ girls permute $$3!$$ ways.
By product rule, the total number is $$3\times 2\times 2!\times 3! =72$$.