Question

In: Advanced Math

Compute the number of ways to deal each of the following five-card hands in poker. (a)...

Compute the number of ways to deal each of the following five-card hands in poker. (a) At least one card from each suit. (b) At least one card from each suit, but no two values matching.

Solutions

Expert Solution

(a) Note that we can get one card from a particular suit (say spades) in 13 ways.

We can think of this problem as we have 5 boxes to fill with each box with a single card and after filling the boxes we must have a card from each suit atleast. We can fill first four boxes one from each suit, that is in ways, and the other box with any card except those four.

Now after filling first four boxes we have cards remaining. So the last box can consists any card from the remaining 48, hence it can be filled in 48 ways. Hence the total number of way one can fill 5 boxes such that each box contains atleast one card from each suit in many ways.

Note that in this the ordering of receiving cards is immaterial.

(b) Same way we'll think. We have to fill five boxes with a single card and after filling the boxes we must have a card from each suit atleast and no two cards match.

Note that for the first box we can fill it by 13 ways.

For the second box we choose a different suit and a different number than the first one. In a suit we have 13 cards, for the second box we can choose any one from these 13 cards except one, which got the same number as the first one. So we can fill in 13-1=12 ways.

For the third box we can fill in from any one card from another suit not having numbers as the first one as well as the second one, that is in 13-2=11 ways.

And by the same argument for the fourth box we can choose in 13-3=10 ways.

So at this point we got four boxes with 4 different numbered cards from four different suit. So remaining cards are 52-4=48. Now suppose the number of the card in the first box is a, second box is b, third box is c and the fourth box is d. So there are still 3 many cards for each numbered a, b, c, d remaining. We must exclude these cards to from our choice to get no repetition. Hence for the fifth box we got many ways. Hence the total number of way is


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