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ybengal33 · Feb 25, 2009, 03:14 PM

If a couple has four boys and four girls, how many different gender sequences are possible?:confused:

Perito · Feb 25, 2009, 04:02 PM

I was wrong.

ebaines · Feb 26, 2009, 11:14 AM

Ithink it's:

$<br /> \frac {8!} {4! \cdot 4!} \ = \ 70<br />$

Here's why:

There are 8! Unique ways the 8 children can be arranged. Since we don't care about the order of the 4 boys, we divide by the numer of ways 4 boys can be arranged, or 4! That is, we don't care whether the order is Tom, Harry, and Sam or Tom, Sam and Harry - so we divided by the number of ways you can arange 4 boys. Similarly we don't care how the girls are arranged, so we divide again by 4! That leaves the number of ways 8 children consisting of 4 boys and 4 girls can be arranged if you all you care about is gender.

It can help to check one's work for probems like this if you use smaller numbers in the formula and then write out all combinations to see if they agree. For example, if there were three boys and three girls my fomula gives 6!/(3! * 3!) = 20. And indeed there are 20 ways to arrange 3 boys and 3 girls.