allabout wrote:

@ BG

According to your logic you need to multiply 13C2 with 4, since there are four suits a 13 cards.

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@ ywilfired

I can't evaluate why 13*13*4C2 should be wrong.

13*13 means the 13^2 ways to combine the cards two suits.

4C2 means the number of combinations you can choose two suits out of four.

What is wrong with it?

Allow me to clarify. What I have learnt here is

1. If question asks # of ways then the order matters.
i.e 52*39

2. If question asks how many different combinations then order doesn't matter.
i.e 13C2 * 13C2 * 4C2

Lets prove this by taking 8 cards of four suites. Say A1, A2, B1, B2,C1,C2 and D1,D2

For case 1 above:
(A1, another card other than A2) = 6 ways

(A2, another card other than A1) = 6 ways

(B1, another card other than A1,A2,B2) = 4 ways

(B2, another card other than A1,A2,B1) = 4 ways

(C1, another card other than A1,A2,B1,B2,C2) = 2 ways

(C2, another card other than A1,A2,B1,B2,C1) = 2 ways

Total = 24.

But there is difference between (A1,B1 - First A1 drawn and then B1 drawn) and (B1.A1 - First B1 drawn and then A1 drawn) if # of ways are asked.
So total = 24 * 2 = 48 i.e 8* 6 (Equivalent to 52*39)

For case 2 above:
(A1, another card other than A2) = 6 ways

(A2, another card other than A1) = 6 ways

(B1, another card other than A1,A2,B2) = 4 ways

(B2, another card other than A1,A2,B1) = 4 ways

(C1, another card other than A1,A2,B1,B2,C2) = 2 ways

(C2, another card other than A1,A2,B1,B2,C1) = 2 ways

Total = 24.

There is no difference between (A1,B1) and (B1.A1) if # of diferent combinations are asked.
So total = 24 i.e 2 * 2 * 4C2 (Equivalent to 13 * 13 * 4C2)

Hope this clears all doubts. I also calculated wrong but good thing is I learned a new concept.

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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008