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

1 Answer
Feb 28, 2018

Answer:

#""^6C_3 = 20#

Explanation:

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

#""^nC_0, ""^nC_1, ""^nC_2, ..., ""^nC_(n-1), ""^nC_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 ]]

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So we find #""^6C_3 = 20#