Question 1106806
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The other tutor overlooked one solution....<br>
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.<br>
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).<br>
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).<br>
So there are three solutions to P(n,k) = 120: P(6,3), P(5,4), and P(5,5).