Expert: Josh - 3/17/2005

Question
we deal randomlu 5 cards from a deck of 52. two deals differing only by the order are considered the same.
a) how many different deals are there?
b)how many of them cointain exactly two aces?
c)how many contain at least one card of a red suit and at least one card of a black suit.

i am finding it quite difficult to comprehend and complete this question. it would be very much appreciated if you could tell me step by step how to complete it. please, this is urgent.
thankyou.

Answer
You start with a deck of 52 cards.

Pick one in random and give it to person A.
Now, you are left with 51 cards.

Pick the second one in random from 51 cards, there are 51 possibilities. Afterwards, you have 50 cards.

Pick the third one in random from 50 cards, there are 50 possibilities. It then leaves you with 49 cards.

Get the idea?
This leads to 52*51*50*49*48 possible permutations, if the order matters not, we must divide this by 5!=120 to account for the different arrangements.

To see this, suppose you have chosen 10,J,Q,K,A just for the purpose of discussion here.

10,J,Q,K,A
10,J,Q,A,K
...
J,Q,K,A,10
...
A,K,Q,J,10 etc (25 arrangements in total) are all regarded the same.

General formula for part (a) is as follows.
Number of combinations, C(N,j)=N!/[j!*(N-j)!]

where factorial n!=n*(n-1)*(n-2)*...*2*1 is the product of integers from n, all the way down to 1. for instance, 4!=4*3*2*1=24. By definition, 1! and 0! both equal to 1.

Interpretation - this formula gives number of possible combinations, when you select j cards from a deck N in random. Important thing is that the order doesn't matter, we treat them to be the same.

b) You need two cards from {spade(A),heart(A),club(A),diamond(A)}. Again, if order matters not, this gives rise to C(M,k) combinations, you must pick two from this set of four. In this case, M=4, k=2.
C(4,2)=4!/[(4-2)!*2!]=24/[2*2]=6 possible combinations.
Additionally, the remaining three cards must be drawn from the 48 cards (excluding the aces). The number of combinations here is C(48,3).
Overall, there are C(4,2)*C(48,3) combinations.

c) This is same as saying, we cannot have them all red or all black. 26 cards are black and 26 are red.
To pick 5 blacks, there are C(26,5) ways.
To pick 5 reds, similarly, there are C(26,5) ways.
What we want is neither of these.
So, the answer is C(52,5)-[2*C(26,5)].

Cheers