# In a bridge game of playing cards,4 players are distributed one card each by turn so that each player gets 13 cards. Find possible number ways?

If we care about the order of the deal, 52!~=8.07xx10^67

If we don't care about the order of the deal, (52!)/(13!)^4~=5.4xx10^28

#### Explanation:

The question talks about the way the cards are dealt, and so I'll answer this question as a permutation (where order matters).

The first card dealt can be any of the 52 cards of the deck.
The second card dealt can be any of the remaining 51 cards.
The third card dealt can by any of the remaining of the 50 cards.
And so on.

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 $n = \text{population", k="picks}$

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:

${C}_{52 , 13} \times {C}_{39 , 13} \times {C}_{26 , 13} \times {C}_{13 , 13}$

(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