How many ordered pairs of positive integers 1≤k≤n≤50 are there, such that k divides n, and (n/k )!= n!/k! ?
To solve this problem, we need to find the number of ordered pairs of positive integers (k, n) that satisfy the given conditions.
Let's break this down step by step:
Step 1: Simplify the equation
Since (n/k)! is equal to (n! / k!), we can simplify the equation as follows:
(n/k)! = n!/k!
Step 2: Simplify the factorials
To simplify the factorials, we can use the property of factorials:
(n/k)! = (n! / k!) = n! / (k! * (n-k)!)
Step 3: Rewrite the equation using the simplified factorials
Now, let's rewrite the equation using the simplified factorials:
n! / (k! * (n-k)!) = n!/k!
Step 4: Cancel out common factors
Since the numerator and denominator of the equation both have n!, we can cancel it out:
1 / (k! * (n-k)!) = 1/k!
Step 5: Simplifying further
To simplify further, multiply both sides of the equation by k!:
1 = (n-k)! / (k!(k-1)!)
Step 6: Analyzing the equation
Since both sides of the equation are integers, the only possible value for (n-k)! / (k!(k-1)!) to make the equation true is when (n-k)! is equal to (k!(k-1)!).
Step 7: Substituting values
Now, let's substitute values for k and (n-k) to find the number of ordered pairs (k, n) that satisfy the conditions:
a) If k = 1:
(n-1)! = 1 * (0!)
This implies that (n-1)! = 1, which is true only for n = 2.
b) If k = 2:
(n-2)! = 2! * (1!)
This implies that (n-2)! = 2, which is true only for n = 3.
From the analysis above, we can see that the only possible values for k and n are (1, 2) and (2, 3). Therefore, there are only 2 ordered pairs of positive integers (k, n) that satisfy the given conditions: (1, 2) and (2, 3).