# How do you evaluate ""^6C_3 using Pascal's triangle?

Feb 28, 2018

""^6C_3 = 20

#### Explanation:

The $n + 1$th row of Pascal's triangle consists of the binomial coefficients:

${\text{^nC_0, ""^nC_1, ""^nC_2, ..., ""^nC_(n-1), }}^{n} {C}_{n}$

So ""^6C_3 is the fourth (middle) term of the $7$th row of Pascal's triangle - the row that begins $1 , 6$.

[[ Note: Some authors call the first row of Pascal's triangle the $0$th row, in which case this would be called the $6$th row ]]

So we find ""^6C_3 = 20