TAOCP 1.2.6 Exercise 4
By Exercise 3, the number of bridge hands is \binom{52}{13} = \frac{52!
Section 1.2.6: Binomial Coefficients
Exercise 4. [10] Give the answer to exercise 3 as a product of prime numbers.
Verified: no
Solve time: 29s
By Exercise 3, the number of bridge hands is
$$ \binom{52}{13} = \frac{52!}{13!,39!} = \frac{52\cdot51\cdot50\cdots40}{13!}. $$
Factoring each term into primes and canceling the factors of $13!$, we obtain
$$ \binom{52}{13} = 2^2\cdot3^3\cdot5\cdot7\cdot11\cdot17\cdot19\cdot23\cdot41\cdot43\cdot47. $$
Hence the number of bridge hands, expressed as a product of prime numbers, is
$$ \boxed{2^2\cdot3^3\cdot5\cdot7\cdot11\cdot17\cdot19\cdot23\cdot41\cdot43\cdot47}. $$