For how many pairs of positive integers n and k with n and k less than or equal to 20, is the number (2n)!(2k)!n!k!(n+k)! an integer?
To find the number of pairs of positive integers (n, k) for which the given expression is an integer, we need to consider the factors in the expression and their exponents.
Given expression: (2n)!(2k)!n!k!(n+k)!
Let's break it down:
1. (2n)! represents the factorial of 2n.
2. (2k)! represents the factorial of 2k.
3. n! represents the factorial of n.
4. k! represents the factorial of k.
5. (n+k)! represents the factorial of (n+k).
To determine if the expression is an integer, we need to ensure that the exponents of prime numbers in the numerator (factorials) are greater than or equal to the exponents of the same prime numbers in the denominator.
Step 1: Prime factorization of the expression
Let's examine the prime factors:
a) 2: The exponents of 2 in (2n)!, (2k)!, and (n+k)! will be greater than or equal to the exponents in n!, k!, and (n+k)!.
b) Other primes (3, 5, 7, 11, 13, 17, 19, 23...): Since n and k are less than or equal to 20, and (n+k) is also less than or equal to 40, the exponents of these primes in the numerator will always be greater than the exponents in the denominator.
Step 2: Counting pairs (n, k) that satisfy the condition
Given the above analysis, we can conclude that any positive integer values of (n, k), where n and k are less than or equal to 20, will make the expression an integer. Therefore, the number of pairs (n, k) that satisfy the condition is the total number of combinations possible, which is (20C2) = 190.
Hence, there are 190 pairs of positive integers (n, k) with n and k less than or equal to 20 for which the expression (2n)!(2k)!n!k!(n+k)! is an integer.