**joshua_green** | Aug. 5th, 2009 09:20 pm Playing with Chips Suppose you have a stack of *N* poker chips. If *N* ≥ 2, you split the stack into two smaller stacks of sizes *m* and *n* with both new sizes positive integers and, of course, *m* + *n* = *N*. While you're at it, note the product *mn*. You can now repeat the process, splitting any stack(s) of size ≥ 2 into two smaller stacks and noting the product of the two new sizes. After *N* - 1 steps, you will have *N* stacks of size 1 each, hence no more splitting will be possible, and you will have noted a bunch of products. Add all these products together. As an example, if *N* = 5, you could follow the final steps: - Split the stack of size 5 into stacks of size 2 and 3 and note the product 2 × 3 = 6.
- Split the stack of size 2 into stacks of size 1 and 1 and note the product 1 × 1 = 1.
- Split the stack of size 3 into stacks of size 2 and 1 and note the product 2 × 1 = 2.
- Split the stack of size 2 into stacks of size 1 and 1 and note the product 1 × 1 = 1.
- Add all the products together, arriving at 6 + 1 + 2 + 1 =
**10**. Of course, there are choices along the way, as there may be more than one way to split a stack into two smaller stacks. Show, however, that the final sum of the products (10 in the example above) is independent of the choices made (though it may depend on *N*).**Current Mood:** amused
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