Question

Statistics? Anyone?
I'm not sure if I'm getting this question right.

A family has nine children. If this family has exactly five boys​, how many different birth and gender orders are​ possible?

Answer 1

I'm coming out to:\[\frac{ 9! }{ 5!*4! }\]

Answer 2

I haven't used LaTeX in a while, it'll take me sometime to write it out manually

Answer 3

First assign gender, then assign positions.

Answer 4

Oh wait, never mind, they've already assigned gender, you just need to order them.

Answer 5

Latex isn't working for me, I'll just write it out:

9_C_5 = (9!)/(9-5)!5!

Answer 6

Yeah, this a combinations. You choose \(5\) positions from \(1\) to \(9\) to assign the boys, then the girls will fall in place. Thus it is \({9 \choose 5}=\frac{9!}{5!(9-5)!} = \frac{9!}{5!\cdot 4!}\)

Answer 7

Simple arithmetic now,

(362,880)/(120)(24) --> 126

Answer 8

\[
\frac{9!}{5!4!} = \frac{9\cdot 8\cdot 7\cdot 6\cdot \cancel{5!}}{\cancel{5!}4!} = \frac{9\cdot 8\cdot 7\cdot \cancel 6}{4\cdot \cancel{3\cdot 2}} = \frac{9
\cdot 8
\cdot 7}{4} = 9\cdot 2\cdot 7=126
\]

Answer 9

Ahh, thanks! It kinda makes sense. Not all of it, but I'm kinda understanding a bit.