Find the third term of (a-2b)^13

Question

I'm looking a huge Pascal's triangle but I am trying to understand the easier way to do this

Answer 1

(a-2b)^13 =

a^13 + 13a^12(-2b) + 13!/11!2!)a^11(-2b)^2 + ... + 1!3!/r!(13-r)a^r(-2b)^n-r + ... (-2b)^13

the 3rd term is 78a^11(4b^2) = 312a^11b^2

the exponent of a decreases by 2 from 13 to 11, while the exponent of (-2b) increases from 0 to 2

the coefficient of the 3rd term is the number of combinations of 13 taken 11 at a time, or (n/r) = (13/11) = 13!/11!2! = 13(12)/2 = 78. But combine the 78 with the -2^2 from -2b, to get 78x4= 312