Question

The game of bridge is played by $4$players, each of whom is dealt$13$cards. How many bridge deals are possible?

Short answer

$\left(\begin{array}{c}52\\ 13,13,13,13\end{array}\right)=\frac{52!}{(13!{)}^4}$

This result can be obtained either by the multinomial coefficients or by the basic principle.

Step-by-step solution

Given Information.

The game of bridge is played by $4$players, each of whom is dealt$13$cards.

Explanation.

In how many ways can we divide $52$different elements (cards) into $4$groups (hands)$13$.

Consider the player's differences.

The first player can receive any combination$13$of$52$cards, the order of the cards does not matter.

in$\left(\begin{array}{l}52\\ 13\end{array}\right)$ways.

Whatever cards the first player gets, the second player can receive any combination$13out$$(52-13)=39$of the remaining cards.

in $\left(\begin{array}{l}39\\ 13\end{array}\right)$ways.

Explanation.

The third player can have his/hers cards dealt with him

in$\left(\begin{array}{l}26\\ 13\end{array}\right)$ways.

The remaining cards go to the player no.$4$in $\left(\begin{array}{l}13\\ 13\end{array}\right)=1$a possible way. By the basic principle of counting, the final result is:

$\left(\begin{array}{l}52\\ 13\end{array}\right)\left(\begin{array}{l}39\\ 13\end{array}\right)\left(\begin{array}{l}26\\ 13\end{array}\right)=\frac{52·51·\dots ·41·40}{13!}·\frac{39·38·\dots ·28·27}{13!}\frac{26·25·\dots ·15·14}{13!}=\frac{52!}{(13!{)}^4}$

This result can also be obtained by seeing that this is a division of$n=52$different objects into $4$groups$13$. This is the definition of the multinomial coefficients:

$\left(\begin{array}{c}n\\ n_1,n_2,\dots ,n_r\end{array}\right)=\frac{n!}{n_1!n_2!\dots n_r!}n=n_1+n_2+\dots +n_r$.

Explanation.

$\left(\begin{array}{c}52\\ 13,13,13,13\end{array}\right)=\frac{52!}{(13!{)}^4}$

This result can be obtained either by the multinomial coefficients or by the basic principle.