Question 609327
Hi, there--
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There are (at least) to ways you can think about this. First, we have a formula for nCr.

nCr = n! / (r!(n-r)!)
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Since n=r, we can substitute n for r in the formula.
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nCn = n! / (n!(n-n)!)
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Simplify.
nCn = n! / (n! 0!)
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n! over n! is 1. By definition 0! is also 1. So nCn = 1 when n=r.
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Another way to think about this is to remember what combinations mean. How many ways can I choose  r objects from a group of n objects. (Order does not matter.) When n=r, I'm choosing all the objects. Order does not matter, so there is only one way to choose.
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I hope this helps. Feel free to email if you still have questions about this.
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Ms.Figgy
math.in.the.vortex