The particular problem is: how many poker hands (out of ${52 \choose 5}$) result in one pair and three unique cards?
My reasoning goes like this: ${13 \choose 1}$${4 \choose 2}$ gives the amount of pairs (13 values choose 1, 4 suits choose 2 of that suit)
Multiplied by:
${12 \choose 1}$${4 \choose 1}$ (12 remaining values to choose from, 4 suits choose 1)
Multiplied by:
${11 \choose 1}$${4 \choose 1}$ * ${10 \choose 1}$${4 \choose 1}$ by the same logic.
However, the answer is supposed to be:
${13 \choose 1}$${4 \choose 2}$ * ${12 \choose 3}$${4 \choose 1}$${^3}$
and i fail to see why the result should be any different.
Thanks for any insight!