If n ≥ 1 and p is a prime, prove that
(a) (2n)!/(n!)2 is an even integer.
[Hint: Use Theorem 6.10.]
(b) The exponent of the highest power of p that divides (2n)!/(n!)2 is
(c) In the prime factorization of (2n)!/(n!)2 the exponent of any prime p such that n < p < 2n is equal to 1.
We need at least 10 more requests to produce the solution.
0 / 10 have requested this problem solution
The more requests, the faster the answer.