SOLUTION: If P(n,k)=120. Find the possible values of n and k. How many solutions are there?

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Question 1106806: If P(n,k)=120. Find the possible values of n and k. How many solutions are there?
Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
P%28n%2Ck%29=n%21%2F%28n-k%29%21=120

120 = 2×60 = 2×3×20 = 2×3×4×5 = P(5,4) = 1×2×3×4×5 = P(5,5)

Two solutions:

n=5, k=4

n=5, k=5

Edwin

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The other tutor overlooked one solution....

The values of P(n,k) are products of consecutive positive integers. So we are looking for all ways we can get a product of 120 with consecutive positive integers.

7 is not a factor of 120, so the largest positive integer we can use is 6. And if we try using it, we find 6*5*4 = 120. Three consecutive positive integers with the largest being 6 is P(6,3).

What if we start with 5 as the largest positive integers? We find 5*4*3*2 = 120; and then of course 5*4*3*2*1 is again 120. These solutions correspond to P(5,4) and P(5,5).

So there are three solutions to P(n,k) = 120: P(6,3), P(5,4), and P(5,5).