Permutation / Computation Guide

_337_5
Thu, 01 Sep 2005 22:14:51 +0000

For your example: The number 2 can’t be on the ends so: nPr(4,2) would be the number of ways that the remaining 4 numbers can be put on the ends. Then for the middle three, there are 3! or 6 ways to arrange the remaing 3 numbers. Thus, there are 6*nPr(4,2)= 72 ways. Look at all problems in a semi-similar way to what i did. just think, what is the problem asking and how can it be simplified and expressed mathematically?