Question

In how many ways $13$ cards to each $4$ players to be distributed from a pack of $52$ cards and that each may have
A) Ace, King, Queen, Jack of same suit
B) Ace, King, Queen, Jack of any suit

Answer 1

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 queen
For second player, we chose in the same way and similarly for third and fourth player
So 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 $48!$.