combinations

**buckaroobill** · December 16th 2006, 12:52 PM

[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

**ThePerfectHacker** · December 16th 2006, 01:39 PM

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 $n$ ,

$\frac{n!}{4!(n-4)!}+\frac{2n!}{5!(n-5)!}+\frac{n!}{6!(n-6)!}$
Common denominator is,
$6!(n-4)!$
Thus,
$\frac{6\cdot 5 \cdot n!}{6!(n-4)!}+\frac{2\cdot 6\cdot n!(n-4)}{6!(n-4)!}+\frac{n!(n-5)(n-4)}{6!(n-4)!}$
Now you can combine fractions

**Soroban** · December 16th 2006, 03:10 PM

Hello, buckaroobill!

My approach is slightly different . . .

Simplify: . $\binom{n}{4} + 2\binom{n}{5} + \binom{n}{6}$

We have: . $\frac{n!}{4!(n-4)!} + 2\frac{n!}{5!(n-5)!} + \frac{n!}{6!(n-6)!}$

. . $= \;\frac{n(n-1)(n-2)(n-3)}{4!} + \frac{2n(n-1)(n-2)(n-3)(n-4)}{5!}$ $+ \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{6!}$

Factor: . $n(n-1)(n-2)(n-3)\left[\frac{1}{4!} + \frac{2(n-4)}{5!} + \frac{(n-4)(n-5)}{6!}\right]$

Common denominator: . $n(n-1)(n-2)(n-3)\left[\frac{1}{4!}\!\cdot\!\frac{5\!\cdot\!6}{5\!\cdot\! 6} + \frac{2(n-4)}{5!}\!\cdot\!\frac{6}{6} + \frac{(n-4)(n-5)}{6!}\right]$

. . $= \;\frac{n(n-1)(n-2)(n-3)}{6!}\left[30 + 12(n-4) + (n-4)(n-5)\right]$

. . $= \;\frac{n(n-1)(n-2)(n-3)}{6!}\left(n^2 + 3n + 2\right)$ $\;=\;\frac{n(n-1)(n-2)(n-3)}{6!}(n+1)(n+2)$

. . $= \;\frac{(n+2)(n+1)n(n-1)(n-2)(n-3)}{6!} \;=\;\boxed{\binom{n+2}{6}}$

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