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chrisann_julia · Feb 7, 2007, 05:47 AM

How many ways are there to get a 12 with 3 dices?

Capuchin · Feb 7, 2007, 06:38 AM

Firstly, the plural of die is dice. Dices means nothing.

Now the grammar police have come and gone, this isn't so hard to ennumerate yourself.

Why don't you have a go writing them out, and we'll help you to spot any you have missed.

galactus · Feb 7, 2007, 09:32 AM

Expand this out and the coefficient of x^12 is your answer.

$\left(\sum_{k=1}^{6}x^{k}\right)^{3}$

asterisk_man · Feb 8, 2007, 11:55 AM

Based on your question galactus's answer (while totally awesome :) ) is probably over your head. Based on "Dices" I'm guessing capuchin's "enumerate" is also going to be more than you can handle. Try this:
Write out all the different ways that you can roll 3 dice into a table and include the sum. It would start something like this:
1 1 1 = 3
1 1 2 = 4
1 1 3 = 5

then count up all the 12s you have in the right hand column.

for example, how many ways can you get a 3 with 2 dice?

Code:

1 1 = 2    2 1 = 3    3 1 = 4
1 2 = 3    2 2 = 4    3 2 = 5
1 3 = 4    2 3 = 5    3 3 = 6
1 4 = 5    2 4 = 6    3 4 = 7
1 5 = 6    2 5 = 7    3 5 = 8
1 6 = 7    2 6 = 8    3 6 = 9

4 1 = 5    5 1 = 6    6 1 = 7
4 2 = 6    5 2 = 7    6 2 = 8
4 3 = 7    5 3 = 8    6 3 = 9
4 4 = 8    5 4 = 9    6 4 = 10
4 5 = 9    5 5 = 10   6 5 = 11
4 6 = 10   5 6 = 11   6 6 = 12

you can see there are only 2 combinations that equal 3 so the answer is 2.