[note...n choose k entails the following: n!/(k!(n-k)!) i have to simplify following statement but don't know how to go about it: (n choose 4) + 2(n choose 5) + (n choose 6) so if anyone could help me, that would be great
Last edited by buckaroobill; Dec 16th 2006 at 02:20 PM.
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Originally Posted by buckaroobill [note...n choose k entails the following: n!/(k!(n-k)!) i have to simplify following statement but don't know how to go about it: (n choose 4) + 2(n choose 5) + (n choose 6) so if anyone could help me, that would be great For sufficently large , Common denominator is, Thus, Now you can combine fractions
Hello, buckaroobill! My approach is slightly different . . . Simplify: . We have: . . . Factor: . Common denominator: . . . . . . .
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