SOLUTION: If nPr = 336 and nCr = 56, determine what n and r are.

Question 535802: If nPr = 336 and nCr = 56, determine what n and r are.
Answer by fcabanski(1391) (Show Source): You can put this solution on YOUR website!
Look at the formula for each:

nPr=n!/(n-r)! = 336 and nCr=n!/r!(n-r)! = 56

That means 336/r! = 56 and r! = 336/56 = 6 = 3*2 * 1 so r = 3.

n!/(n-3)! = 336 = n*(n-1)*(n-2) (Remember, if the denominator is (n-3)!, it cancels all but the first 3 factors of n!.

Example 8!/(8-3)! = 8*7*6*5*4*3*2*1/5*4*3*2*1= 8*7*6

The easiest way to finish is to try some numbers.

10: 10*9*8 = 720 (too big)

9: 9*8*7 = 504 (too big)

8: 8*7*6 = 336 (That's it)

n=8 and r = 3
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