Ten kids line up for recess. There are five boys and five girls. How many ways are there for the 10 kids to line up so that all the boys are ahead of all the girls?

14400

Step-by-step explanation:

We have 2 groups of 5 elements each

there is a condition all boys will be ahead always, therefore in the group of boys there are

5! = 5*4*3*2*1 = 120 differents form

And the girls at the same time can change theirs places according to

5! = 120

Then total ways for the line up in the conditions of the statement of the problem are

T = 120*120

14400