Question 1163219
If nPr = 336 and nCr = 56, find n and r. 

Please note that:
P and C are permutation and combination respectively. 
<pre>{{{matrix(2,3, "" [n]P[r], "=", 336, " " [n]C[r], "=", 56)}}}

{{{matrix(1,3, " " [n]C[r], "=", " "[n]P[r]/r!)}}}
{{{matrix(1,3, 56, "=", 336/r!)}}} ------ Substituting for {{{matrix(2,3, 56, for, " "[n]C[r], 336, for, " "[n]P[r])}}}  
{{{matrix(1,3, 56r!, "=", 336)}}} ------ Cross-multiplying
{{{matrix(2,3, r!, "=", 336/56, r!, "=", 6)}}} 
{{{highlight_green(matrix(1,3, r, "=", 3))}}}, since 3(2)(1) = 6

With r = 3, we can say that: {{{matrix(3,3, " " [n]P[r], "=", 336, " "[n]P[3], "=", 336, n(n  -  1)(n  -  2), "=", 336)}}}

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

Therefore, we get: {{{highlight_green(matrix(1,7, " "[8]P[3], "=", "336,", and, n, "=", 8))}}}</pre>