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The number of ways of getting three of a kind $(x, x, x, y, z)$ when rolling 5 dice is -

$${5 \choose 3}{2 \choose 1}{6 \choose 1}{5 \choose 2}$$

${5 \choose 3}$ - Ways of choosing dice for the $x$'s

${2 \choose 1}$ - Ways of choosing dice for the $y$

${6 \choose 1}$ - Ways of choosing the number for the $x$'s

${5 \choose 2}$ - Ways of choosing the numbers for the $y$ and $z$

I am looking for the intuition behind why we can't have $6 \choose 3$ for the ways of choosing the numbers for $x, y, z$. I actually know why it is but I could really use a clear intuitive explanation so my head isn't wrecked with it. I want to be able to visualize it quickly and clearly.

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You would still need to choose which of your $3$ number is the tri one –  Jean-Sébastien Nov 11 '12 at 19:24
    
I have a blog article about this that you might find helpful. –  MJD Nov 16 '12 at 23:45

1 Answer 1

Choosing $6\choose 3$ does not specify which value you set the trio with. It requires you to choose $1$ of these for the trio, which can be done in $3\choose 1$. You get $$ {3\choose 1}{6\choose 3}=\frac{3!\,6!}{1!\,2!\,3!\,3!}=\frac{6!}{1!\,2!\,3!}=\frac{6\times 5!}{1!\,2!\,3!}=\frac{6!\, 5!}{1!\,5!\,2!\,3!}{6\choose1}{5\choose2} $$ So they would be both equivalent, if you add the choice for the trio.

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