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janvdl · #2 September 5th, 2007, 05:39 AM

Quote:

Originally Posted by ThePerfectHacker

2. Given a positive integer $n$ define a $k$ -partition to be a sum of $k$ positive integers which sum to $n$ . For example, $n=10$ . The following are $4$ -partitions. $10 = 4+4+2+2$ and $10 = 2+2+4+4$ and $10 = 1+1+1+7$ . Notice that $2+2+4+4\mbox{ and }4+4+2+2$ are considered distinct*. Say you a given a specific $n$ . And given a specific value of $k$ , can you find the total number of $k$ -partitions of this integer, with a formula?** Now try to see how many partitions (again not counting order) exist for a given integer $n$ (the answer is really supprising).

Hint: Review your Combinatorics formula for this one.

$\frac{(n + k - 1)!}{k!(n - 1)!}$

So if $n = 10$ and using $4$ -partitions(including repetitions), then we will have $715$ possibilities.

$\frac{(10 + 4 - 1)!}{4!((10 - 1)!)} = 715$

Am i even close to correct?