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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.
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: