[Math] Number of ways to arrange 3 boys and 5 girls in line if the boys must stand next to each other

combinationspermutations

I'm having difficulty with the following question:

In how many ways can three boys and five girls stand in a straight line, if the boys must stand next to each other?

It says that the answer is $3!$ x $5!$ x $6$ – The six representing the six places for boys. How did they get that number?

Isn't there another way to do it, by using combinations and treating the 'three boys' as ONE group?

Thanks!

Best Answer

Your idea of grouping the boys in one group is entirely sound. There are 6 ways to place that group of boys in the line, $5!=120$ ways of arranging the girls after that and $3!=6$ ways of arranging the boys within their cluster, so the answer must be $6\times5!\times3!=4320$ (which is exactly the same form as the given answer).

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