How many different ways are there to deal four 13 card hands to four players from a standard deck of 52 cards?

Question

This question is from the third Combination example from lecture notes 3A. Researching this question I found this answer

https://www.quora.com/In-how-many-ways-can-52-cards-be-distributed-to-four-players-with-13-cards-each 

What I don't understand is why you multiple the number of ways for each hand as opposed to adding. I know for player 1 it's C(52,13), player 2 C(39,13), player 3 C(26,13), then player 4 is 1 or C(13,13). I thought adding these together would get the total number of ways to deal the cards. Is this answer from the link correct and why? 

 

Answer 1

Instead of giving you the final answer, I ask you simplify question by an example you can solve on paper. Let's ask in how many ways it is possible to distribute 4 cards (A,B,C,D) between two people equally. We can write possible ways as follows:

$|\{(AB, CD), (AC,BD),(AD,BC),(BC,AD),(BD,AC),(CD,AB)\}| = 6$

In the same way in math, I should multiply the ways I can select first 2 cards from 4 cards for the first person, $\binom{4}{2}$, and the remaining 2 cards for the second person $\binom{4-2}{2} = \binom{2}{2}$.   Therefore, $\binom{4}{2} \times \binom{2}{2} = 6$ ways are available.

Now, can you solve the main problem?

Comments

Yes thanks. I just learnt about the Basic Counting Rule(BCR) from the textbook too which made this even more clearer. BCR says that if there are r actions with the first action having m1 possibilities and second having m2 possibilities and so on, then the total number of possibilities is m1*m2... *mr. So with your simplified example,  the number of actions (r) is 2 for disturbing to 2 people
For the first person or action there are 6 possibilities (m1) for distributing 4 cards
For the second person or action  there is only 1 possibility (m2) for distributing the remain 2 cards.

By the BCR the total number of possibilities is(m1*m2) 6*1 which is 6.

This applied to my original question means the answer I found is correct! Thanks again.