SOLUTION: Please help me solve this problem four children, identical twins, and identical triplets pose for a photograph. How many photographs can be made?

Question 1191507: Please help me solve this problem four children, identical twins, and identical triplets pose for a photograph. How many photographs can be made?
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

I'm assuming the identical twins are separate from the four children mentioned at first. The same goes for the identical triplets.

There are 4+2+3 = 9 kids in all
There would be 9! = 9*8*7*6*5*4*3*2*1 = 362,880 different permutations or orderings if we can tell the twins and triplets apart.

However, we can't tell the twins apart, so we have to divide by 2 to fix the overcounting concerning the identical twins.
We also have to divide by 3! = 3*2*1 = 6 to account for the triplets we can't tell apart either.
Consider the set {A,B,C}. There are 6 ways to arrange those three items.

Overall, the massive number calculated earlier is too large by a factor of 2*6 = 12
Meaning we have to divide that large figure by 12 to fix the error.

(362,880)/12 = 30,240
Interesting side note: This is equivalent to 2*(9 P 5) though I'm not sure if it's a coincidence or not.

Answer: 30,240

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