Answer  Hint: A) For the first part, firstly, we will choose the suit to distribute to each player and then calculate the number of ways to distribute them. Then we will distribute remaining cards to all of them.B) For the second part, we will choose ace, king, queen and jack to distribute to each player. Then we will distribute remaining cards to all of them. Complete step-by-step answer: A) Number of ways of choosing a suit from $4$ different suits $ = {}^4{C_1}$Number of ways to arrange these 4 different suits $ = 4!$Now, the number of ways to distribute remaining 48 card to 4 groups is$ = \dfrac{{48!}}{{12! \times 12! \times 12! \times 12!}}$So total number of ways to distribute according the first condition is$ = {}^4{C_1} \times 4! \times \dfrac{{48!}}{{12! \times 12! \times 12! \times 12!}}$$ = 4 \times 4! \times \dfrac{{48!}}{{12! \times 12! \times 12! \times 12!}}$So our answer is $ = 4 \times 4! \times \dfrac{{48!}}{{12! \times 12! \times 12! \times 12!}}$B) For first player, we chose one of ace, jack, king or queenFor second player, we chose in the same way and similarly for third and fourth playerSo ways for this $ = 4 \times 4 \times 4 \times 4 = {\left( 4 \right)^4}$Number of ways to arrange these distributions to all four players $ = 4!$Now, the number of ways to distribute remaining $48$ card to $4$ groups is$ = \dfrac{{48!}}{{12! \times 12! \times 12! \times 12!}}$So total number of ways to distribute according the second condition is$ = {\left( 4 \right)^4} \times 4! \times \dfrac{{48!}}{{12! \times 12! \times 12! \times 12!}}$So our answer is $ = {\left( 4 \right)^4} \times 4! \times \dfrac{{48!}}{{12! \times 12! \times 12! \times 12!}}$Note: We used permutations to solve above questions. Note that in both cases, the number of ways to distribute the remaining $48$ card to $4$ groups is $ = \dfrac{{48!}}{{12! \times 12! \times 12! \times 12!}}$ . We divided by $12! \times 12! \times 12! \times 12!$ because we have to arrange them in groups. If we were arranging $48$ cards to $48$ groups then we would have divided by $1! \times 1! \times 1! \times 1!$ which is eventually $48!$.
First, if the players are indistinguishable, then both answers should be divided by $4!$. Second, the second approach assumes the following: you shuffle the deck, give the first player the first 13 cards, the second player the next 13 cards, and so on. The duplications are the internal order the cards are given to each player. If you take the same deck and shuffle only the first 13 cards, all players will receive the same cards as before, but you count it as a new deck (there are $52!$ of those). So you divide by these repetitions. Example Suppose you have 4 cards to give to 2 players. Cards are labeled ABCD. The number of ways is $\tfrac{4!}{2!2!}=6$ because AB|CD = BA|CD = BA | DC = AB |CD (these are the $4=2!2!$ reptitions of the event that player 1 has in the end the cards $A$ and $B$ and player 2 the other two. If, players are not distinguishable than you have only 3 options, because you have only 3 ways to split the 4 cards into two decks and it doesn't matter which gets which (you can't tell them apart anyway... )
This means that there is #52!# ways to have the cards dealt out. #52! = 52xx51xx50xx...xx2xx1~=8.07xx10^67# If, however, we don't care about the order in which the cards are dealt, then we can use combinations. The general formula for a combination is: #C_(n,k)=(n!)/((k!)(n-k)!)# with #n="population", k="picks"# We can view this as 13 cards drawn from 52 for the first hand... ... and 13 cards drawn from the remaining 39 cards for the second hand... ... and 13 cards drawn from the remaining 26 cards for the third hand... ... and 13 cards drawn from the remaining 13 cards for the fourth hand. This gives: #C_(52,13)xxC_(39,13)xxC_(26,13)xxC_(13,13)# #(52!)/((13!)(52-13)!)(39!)/((13!)(39-13)!)(26!)/((13!)(26-13)!)(13!)/((13!)(13-13)!)# #(52!)/((13!)(cancel(39!)))(cancel(39!))/((13!)(cancel(26!)))(cancel(26!))/((13!)(13!))(cancel(13!))/((cancel(13!))(0!))# #(52!)/(13!)^4~=5.4xx10^28#
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There is a deck of 52 cards. how many ways are there to distribute 13 cards to each of 4 players?
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