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{n \choose k} = \frac{n \cdot (n-1) \cdots (n-k+1)} | {n \choose k} = \frac{n \cdot (n-1) \cdots (n-k+1)} | ||
{k \cdot (k-1) \cdots 1} = \frac{n!}{k!(n-k)!} \quad \mbox{if}\ 0\leq k\leq n \qquad </math> | {k \cdot (k-1) \cdots 1} = \frac{n!}{k!(n-k)!} \quad \mbox{if}\ 0\leq k\leq n \qquad </math> | ||
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| + | //comment Beau Morrison | ||
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| + | Also remember to prove that: | ||
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| + | <math>{n \choose k} = {n \choose n-k}</math> | ||
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| + | when solving the equality. | ||
Latest revision as of 06:41, 15 October 2008
Here's what I did, and it seemed to work. Let me know if I forgot anything.
Simply expand the original $ Pr(x) = \left( \begin{array}{ccc} n \\ x \end{array} \right)p^{x}(1-p)^{n-x} $.
Now, see if substituting n-x for x and expanding results in the same answer.
Don't forget that: $ {n \choose k} = \frac{n \cdot (n-1) \cdots (n-k+1)} {k \cdot (k-1) \cdots 1} = \frac{n!}{k!(n-k)!} \quad \mbox{if}\ 0\leq k\leq n \qquad $
//comment Beau Morrison
Also remember to prove that:
$ {n \choose k} = {n \choose n-k} $
when solving the equality.
