Answer: 10-Card Hand w/ 2 4-of-a-Kinds - No Pairs/3 of a Kinds

[toothpaste666]

Homework Statement

How many 10-card hands are there chosen from a standard 52-card deck in which there are exactly two 4-of-a-kinds; no pairs or 3-of-a-kinds?

The Attempt at a Solution

if there are exactly 2 4 of a kinds that takes up 8 of the 10 cards in the hand and the remaining 2 must be of different kinds since there are no pairs. first we choose 2 kinds for the 4 of a kinds: 13C2 ways . now we have to make sure the last 2 cards are of different kinds so we pick 2 of the remaining 11 kinds: 11C2 ways. now for each of these kinds there are 4 ways to pick a card from them so 4x4 = 16.

the whole process is:
(13C2)(11C2)(16)

is this correct? I want to make sure I am not double counting anything

 

[krebs]

Looks good; I got the same answer with the same reasoning before looking at your solution.

 

[Samy_A]

By using a slightly different reasoning, I got ##{{13}\choose{2}}({{44}\choose{2}}-11{{4}\choose{2}})##
Both our results give 68640, so that looks promising.

 

[PeroK]

Another way to finish it off:

Once you have the two fours. The next card can be anything (44 cards) and the last card anything of a different kind (40 cards).

 

[krebs]

Another way to finish it off:

Once you have the two fours. The next card can be anything (44 cards) and the last card anything of a different kind (40 cards).

This doesn't work because you would double count. Order of the draw doesn't matter, so you have to divide by 2!.

$$16\frac{11!}{2!9!} = \frac{16*11*10}{2} = \frac{44*40}{2}$$

 

[PeroK]

Yes, that's what I meant!

 

[toothpaste666]

thanks everyone