How many ways are there to distribute five cards each to six players from a pack of 52 cards?

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Answer

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Hint:Use the formula for combination, that is, \[{}^n{C_r} = \dfrac{{n!}}{{r!(n - r!)}}\] to evaluate the number of ways.In First question dividing among four players equally in order, first, distribute 13 cards to player 1, then 13 to player 2, and so on until player 4.In second for forming them into 4 groups of 13 each, the order doesn’t matter here. Then, the 52 cards are divided into 4 sets of 17, 17, 17, and 1.Here also order doesn’t matter.Hence,we get the required answer.

Complete step-by-step answer:

We are given a pack of 52 cards, in the first part of the question, we need to divide them equally among four players in order.We know that 52 divided by 4 is 13, hence, all the players get 13 cards each.First, we give 13 cards out of 52 to player 1, the number of ways is \[{}^{52}{C_{13}}\].Now, we have 39 cards, the number of ways of giving 13 cards to player 2 is \[{}^{39}{C_{13}}\].Now, we have 26 cards, the number of ways of giving 13 cards to player 3 is \[{}^{26}{C_{13}}\].Now, we have 13 cards, the number of ways of giving 13 cards to player 4 is \[{}^{13}{C_{13}}\].Let the total number of ways of distributing 52 cards equally among 4 players in order be \[{N_1}\].\[{N_1} = {}^{52}{C_{13}} \times {}^{39}{C_{13}} \times {}^{26}{C_{13}} \times {}^{13}{C_{13}}\]The formula for combination is as follows:\[{}^n{C_r} = \dfrac{{n!}}{{r!(n - r!)}}............(1)\]Using, formula (1), we have:\[{N_1} = \dfrac{{52!}}{{13!(52 - 13!)}} \times \dfrac{{39!}}{{13!(39 - 13!)}} \times \dfrac{{26!}}{{13!(26 - 13!)}} \times \dfrac{{13!}}{{13!(13 - 13!)}}\]\[{N_1} = \dfrac{{52!}}{{13!.39!}} \times \dfrac{{39!}}{{13!.26!}} \times \dfrac{{26!}}{{13!.13!}} \times \dfrac{{13!}}{{13!.0!}}\]Canceling common terms, we have:\[{N_1} = \dfrac{{52!}}{{{{(13!)}^4}}}\]In the second part of the question, we need to find the number of ways of forming 52 cards into 4 groups with 13 cards each.It is the same as the first question but without order, hence, we divide the first question’s answer by 4! for four identical groups.\[{N_2} = \dfrac{{52!}}{{{{(13!)}^4}(4!)}}\]In the third part of the question, we have to divide 52 cards into 4 sets, three having 17 cards each and the last set having only one card. Let the number of ways of doing so be \[{N_3}\].First, we give 17 cards to the first set, then we have 35 cards, then we give 17 cards to the second set, then we have 18 cards, then we give 17 cards to the third set and the remaining one card to the last set. The order does not matter hence, we divide by \[3!\] for three identical groups.\[{N_1} = \dfrac{{{}^{52}{C_{17}} \times {}^{35}{C_{17}} \times {}^{18}{C_{17}} \times {}^1{C_1}}}{3}\]\[{N_3} = \dfrac{{52!}}{{17!.(52 - 17)!}} \times \dfrac{{35!}}{{17!.(35 - 17)!}} \times \dfrac{{18!}}{{17!.(18 - 17)!}} \times \dfrac{{1!}}{{1!.(1 - 1)!}}\dfrac{1}{{3!}}\]\[{N_3} = \dfrac{{52!}}{{17!.35!}} \times \dfrac{{35!}}{{17!.18!}} \times \dfrac{{18!}}{{17!.1!}} \times \dfrac{{1!}}{{1!.0!}}\dfrac{1}{{3!}}\]\[{N_3} = \dfrac{{52!}}{{{{(17!)}^3}(3!)}}\]Hence, we found the number of required ways.

Note: You might get confused with the second question and think it is the same as the first question, but remember, the order is important in the first question while it is not important in the second question, hence, we need to divide by 4!.

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The answer is $\frac{52!}{(5!\times 5!\times 5!\times 5!\times 32!)} = 1.4782628e+24$

This reminds me of that if I have to select 7 objects from 3 types with infinite supply:

oo|oo|ooo

I know that it's no necessary to use the bar in this question because each person has the same numbers of the cards, but I still want to try to add 4 bars.

That is:

5 cards for person1 | 5 cards for person2 | 5 cards for person3 | 5 cards for person4 | undealt cars

so the equation is: $\frac{56!}{(5!\times 5!\times 5!\times 5!\times 32!\times 4!)}=5.4295116e+29$

What's wrong with my idea?