# Pascal's triangle

Pascal's triangle up to row six

Pascal's triangle is a triangular arrangement of the coefficients of a binomial expansion. Although it was discovered centuries earlier,[1] the triangle's full significance was first understood by 17th century mathematician and religious scholar Blaise Pascal.[2] Pascal's triangle is constructed by starting with a single 1, and then moving downward and out so that each new element is the sum of the two numbers diagonally above it. (Elements outside the triangle are considered to be zero.)

The top row of the triangle (containing just a single 1) is the 0th row, with rows below it being 1st, 2nd, and so on. The left-most element in each row (which will always be a 1) is the 0th element of that row, with the following elements continuing as 1st, 2nd, and so forth.

## Applications

Pascal's triangle has numerous applications and properties, some of which are beyond the scope of this article.[3] Some are useful, others are merely interesting. Only a few of the more well-known applications of Pascal's triangle are described below.

### Binomial Expansion

For the binomial (a + b)n, the coefficients of the expansion are found in the nth row of Pascal's triangle. (Remember the top row is row 0.) For example, the binomial coefficients for (a + b)4 are the numbers in row 4 of the triangle. (a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4

### Combinations

For a 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 3rd element of the 5th row. 5C3 = 10

### Pascal's Rule

Pascal's rule states that:

${n \choose r} = {n-1 \choose r-1} + {n-1 \choose r}$

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:

${n-1 \choose r-1} + {n-1 \choose r}$
$= \frac{(n-1)!}{(r-1)!(n-1-(r-1)!}+ \frac{(n-1)!}{r!(n-1-r)!}$
$= \frac{ (n-1)!r + (n-1)!(n-r)}{r!(n-r)!}$
$= \frac{ (n-1)!( r + (n-r)}{r!(n-r)!}$
$= \frac{n!}{r!(n-r)!} = {n \choose r}.$