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1.how many bridge hands contain 3 aces and exactly one two-card suit?

2.how many bridge hands contain 3 aces and exactly one two-of-a-kind?

For the first question,I guess the answer is :C(4,3)C(12,2)C(4,2)C(35,8)

Since the denomination is fixed ,I only need to choose 3 suits for the aces.For the two-card suit,I need to choose 2 numbers excluding A,so I choose 2 denominations out of 13-1=12.There's still 13-3-2=8 cards left.Again,I can't choose A and the same suit as the 2 cards.Then ,the total number of cards I can choose is:52-4-13=35.

For the second question,the answer may be:C(4,3)C(12,1)C(4,1)C(44,8)

In this case,the 2 cards have the same denomination but different suits.So,we still need to choose 3 suits for the aces.But for the 2 cards,we only need to choose 1 number from 12 ,and since there are 2 different suits,we need to choose 2 out of 4.For the remaining 8 cards,we can't choose A again,so is the number in the 2 cards.That counts for:52-4-4=44 cards where the 8 cards derive.

The main difference is the distribution of the 2 card and remaining 8 cards.

Are they correct?

JimfyWinsy
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  • Don't post your question like this - it's ridiculous. The title should be a title. The question bullets (or sections) should be within the body of the question. 2. What do you mean "I guess" and "may be"??? Please include the steps that you've made towards reaching those solutions, so someone can point out what (if anything) is wrong with them.
    • – barak manos May 19 '15 at 06:05
  • Could you explain how you arrived at your answers? – N. F. Taussig May 19 '15 at 11:36
  • @N.F.Taussig The OP updated the question. – layman May 20 '15 at 03:31