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For a [[Combinatorics#Combinations|mathematical combination]] of the form nCr, the solution can be found at row n, element r of Pascal's triangle. For the combination 5C3, the solution is the number at the 3<sup>rd</sup> element of the 5<sup>th</sup> row.
For a [[Combinatorics#Combinations|mathematical combination]] of the form nCr, the solution can be found at row n, element r of Pascal's triangle. For the combination 5C3, the solution is the number at the 3<sup>rd</sup> element of the 5<sup>th</sup> row.
<math>5C3 = 10</math>
<math>5C3 = 10</math>
===Pascal's Rule===
Pascal's rule states that:
:<math> {n \choose r} = {n-1 \choose r-1} + {n-1 \choose r}</math>
In terms of the triangle, this simply means that an entry is the sum of two diagonal entries above it. It is easy to prove:
:<math> {n-1 \choose r-1} + {n-1 \choose r}</math>
:<math> = \frac{(n-1)!}{(r-1)!(n-1-(r-1)!}+ \frac{(n-1)!}{r!(n-1-r)!}</math>
:<math> = \frac{ (n-1)!r + (n-1)!(n-r)}{r!(n-r)!}</math>
:<math> = \frac{ (n-1)!( r + (n-r)}{r!(n-r)!}</math>
:<math> = \frac{n!}{r!(n-r)!} = {n \choose r}.</math>
==Further Reading==
==Further Reading==