For how many ordered 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?
yr pls someone solve it.....
No-one will solve it for you you cheat!!! Copying questions from Brilliant huh? Lame, lame!!!
so whats ur problem i mean usmey aapko problem kya hai..?
To determine the number of ordered pairs of positive integers (n, k) where ((2n)!(2k)!)/(n!k!(n+k)!) is an integer, we need to analyze the divisibility of the given expression.
Let's break it down step by step:
1. First, consider the expression ((2n)!(2k)!). The exclamation mark represents the factorial operation, which means multiplying all positive integers up to a given number. So, (2n)! means multiplying all positive integers up to (2n), and (2k)! means multiplying all positive integers up to (2k).
2. Now, consider the denominator n!k!(n+k)! in the expression. Again, this represents the factorial operation for n, k, and (n+k).
To ensure that the given expression is an integer, the numerator must have at least the same number of prime factors as the denominator, since dividing the numerator by the denominator should not result in any remainders.
Since the question limits n and k from 1 to 20, we can analyze the prime factorizations of each number from 1 to 20.
Prime factorizations of 1 to 20:
1 - none
2 - 2
3 - 3
4 - 2^2
5 - 5
6 - 2 x 3
7 - 7
8 - 2^3
9 - 3^2
10 - 2 x 5
11 - 11
12 - 2^2 x 3
13 - 13
14 - 2 x 7
15 - 3 x 5
16 - 2^4
17 - 17
18 - 2 x 3^2
19 - 19
20 - 2^2 x 5
Now, let's analyze each factor in the numerator ((2n)!(2k)!):
- Every number from 1 to 2n will be included in the numerator, so the prime factors up to 2n will be taken into account.
- Similarly, every number from 1 to 2k will be included, so the prime factors up to 2k will be counted.
Now, let's analyze each factor in the denominator n!k!(n+k)!:
- Every number from 1 to n will be included, so the prime factors up to n will be counted.
- Similarly, every number from 1 to k will be included, so the prime factors up to k will be counted.
- Finally, every number from 1 to (n+k) will be counted, so the prime factors up to (n+k) will also be considered.
To ensure that the expression is an integer, the prime factors from the numerator should cover the prime factors from the denominator. In other words, any prime factor that appears in the denominator should also appear in the numerator with equal or greater exponent.
Based on this analysis, we can conclude that for the given expression to be an integer, the prime factors up to (n+k) should be covered at least once by the prime factors up to 2n and 2k together.
Now, let's consider each prime number and count how many times they are covered by the factors:
- 2: The exponent of 2 in the numerator ((2n)!(2k)!) is (2n + 2k) since each number up to 2n and 2k contributes one factor of 2. In the denominator (n!k!(n+k)!), the exponent of 2 is (n + k). So, for the expression to be an integer, (2n + 2k) should be equal to or greater than (n + k). Simplifying this inequality, we get n + k <= 2n + 2k. Rearranging, k <= n.
- 3: Similarly, the exponent of 3 in the numerator is (2n + 2k) and in the denominator is (n + k). Again, (2n + 2k) should be equal to or greater than (n + k). So, the same inequality holds: k <= n.
- The remaining prime numbers (5, 7, 11, 13, 17, 19) follow the same pattern, and the inequality k <= n applies to them as well.
Since all prime factors require k to be less than or equal to n, the range of possible values for k is limited. By checking all combinations of k and n (k <= n), we can determine the number of ordered pairs (n, k) that satisfy the given condition.
To summarize, to find the number of ordered pairs (n, k) where ((2n)!(2k)!)/(n!k!(n+k)!) is an integer, we need to count the number of valid pairs (n, k) where k <= n for the given range of n and k (1 to 20).