There are 9 children playing in a playground. In a game, they all have to stand in a line such that the youngest child is at the beginning of the line. How many ways can the children be arranged in the line?
a.
40,320
c.
16,777,216
b.
362,880
d.
387,420,489

Question

There are 9 children playing in a playground. In a game, they all have to stand in a line such that the youngest child is at the beginning of the line. How many ways can the children be arranged in the line?
a.
40,320
c.
16,777,216
b.
362,880
d.
387,420,489

oaktree · Answer

OK. There is one option for the first child. The other eight can be arranged in any way, or 8! ways. So there are 1*8!, or 8!, or 40320 ways.

anonymous · Answer

To figure this out, you start out with 9 possibilities for the first spot, 8 for the second, and so on, so the beginning expression we should start with is: 9x8x7x6x5x4x3x2x1.
Now the one kid on the start has only one possibility, so we start out with 1:
1x8x7x6x5x4x3x2x1, which is also 8!, or 8 factoral.