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To answer the following task I can think of two different approaches yet they produce different results. My question is : which way is the right one and why are they different ?

Task : From a standard card game of 52 cards, in how many ways can I get 3 different kinds ?

Approach 1 :

Select 3 different kinds : ${13 \choose 3}$

For each kind choose one card : ${4 \choose 1}^{3}$

Total : ${13 \choose 3}{4 \choose 1}^{3}$

Approach 2 :

Select a first card from all cards : ${52 \choose 1}$

remove the 3 other cards of that kind and pick another card : ${48 \choose 1}$

remove the 3 other cards of that kind and pick the third card : ${44 \choose 1}$

Total : ${52 \choose 1}{48 \choose 1}{44 \choose 1}$

Thanks !

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1 Answer 1

up vote 3 down vote accepted

The reason that they give different results, is the following.

Situation 1:

We pick a collection of three cards of different kinds.

Situation 2:

We pick a sequence of three cards of different kinds.

Namely, in situation 2, we have distinguished between e.g. "pick a Jack, 10, 7" and "pick a 10, Jack 7".

If we correct for this unintended behaviour (by dividing by $3! = 6$, the number of possible sequences of a given collection of three distinct elements) then we obtain precisely the answer from situation 1.

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Ok go it : while the first one handles combinations the second one handles permutations. Thank you @Lord_Farin ! –  Daniel Danny May 1 '13 at 10:32

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