# Binomial Coefficient/Examples/Number of Bridge Hands

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## Contents

## Theorem

The total number $N$ of possible different hands for a game of bridge is:

- $N = \dfrac {52!} {13! \, 39!} = 635 \ 013 \ 559 \ 600$

### Prime Factors

The prime decomposition of the number of bridge hands is given as:

- $\dbinom {52} {13} = 2^4 \times 5^2 \times 7^2 \times 17 \times 23 \times 41 \times 43 \times 47$

## Proof

The total number of cards in a standard deck is $52$.

The number of cards in a single bridge hand is $13$.

Thus $N$ is equal to the number of ways $13$ things can be chosen from $52$.

Thus:

\(\displaystyle N\) | \(=\) | \(\displaystyle \dbinom {52} {23}\) | Cardinality of Set of Subsets | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {52!} {13! \left({52 - 13}\right)!}\) | Definition of Binomial Coefficient | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {52!} {13! \, 39!}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 635 \ 013 \ 559 \ 600\) | after calculation |

$\blacksquare$

## Sources

- 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $3$