SOLUTION: Show that n! /r!(n-r)! + n!/(r-1)! (n-r+1)! = (n+1)! /r!(n-r+1)!

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Question 1135865: Show that n! /r!(n-r)! + n!/(r-1)! (n-r+1)! = (n+1)! /r!(n-r+1)!
Answer by math_helper(2461)   (Show Source): You can put this solution on YOUR website!
Show that n! /r!(n-r)! + n!/(r-1)! (n-r+1)! = (n+1)! /r!(n-r+1)!
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Please be precise, looking at the first term, n! / r!(n-r)! is equal to (n!/r!)*(n-r)! which is probably NOT what you wanted. You undoubtedly wanted n! / (r!(n-r)!). Same for the other terms. Be sure to use parentheses to convey the proper meaning.


Using nCr notation, where nCr = n!/(r!(n-r)!), the problem is to show
nCr + nC(r-1) = (n+1)Cr (1)
Lets say you have n+1 elements: {A1, A2, ... , An, An+1}
Put a star * on any one element.
The RHS of (1) is the selection of r elements from the set of n+1 elements.

One can choose these r elements in two mutually exclusive ways:
1. They can exclude the * element

This can be done in nCr ways (all r elements are selected from the non-starred elements, and there are n of them).
2. They can include the * element

This can be done in nC(r-1) ways (because the * element is pre-selected, this leaves a selection of the remaining r-1 elements from the remaining n elements)

Adding 1. and 2. gives nCr + nC(r-1) = (n+1)Cr

For completeness, writing out the equations:




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