SOLUTION: If nPr = 336 and nCr = 56, find n and r. Please note that: P and C are permutation and combination respectively.

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Question 1163219: If nPr = 336 and nCr = 56, find n and r.
Please note that:
P and C are permutation and combination respectively.

Found 3 solutions by ikleyn, solver91311, MathTherapy:
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.

The ratio of these quantities,  nPr and nCr, is  n*(n-1)* . . . *(n-r+1) :   and equals to r!;


so,  r! =  = 6.


Hence, r = 3.


Next,  nPr at r = 3 is  n*(n-1)*(n-2) = 336.


After several simple trials and errors you get  n = 8.


You may also use  (n-1) ~  = 6.95 (approx.)


ANSWER.  n = 8,  r = 3.

Solved.



Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!




and



Hence







336 divided by 8 is 42. Then 6 and 7 are factors of 42. So the numerator of nPr is 8 X 7 X 6, and then the denominator must be 5! and 8 - 3 is 5, so




John

My calculator said it, I believe it, that settles it


Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
If nPr = 336 and nCr = 56, find n and r.
Please note that:
P and C are permutation and combination respectively.



 ------ Substituting for   
 ------ Cross-multiplying
 
, since 3(2)(1) = 6

With r = 3, we can say that: 

This means that there are 3 CONSECUTIVE, DESCENDING INTEGER-FACTORS of 336. These are: 8, 6, and 7

Therefore, we get: 

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