Question 1191507
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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: <font color=red>30,240</font>
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