Use bionomial formula to find the coefficient of the t^2 m^8 term in the expansion of (3t+m)^10.
I got so far this 10C8 x 3^2 x t^2 x m^8
It tell me the next step is here
=45 x 9 x t^2 x m^8
where did the 45 come from?? It doesn't explain that to me! The answer is 405 by the way but. I don't know how they got to 45!! Please help!!
10C8 = 10!/ [ 8! (10-8)! ]
= [ 10 * 9 ] * 8! / [ 8! * 2! ]
= 90/2
= 45
What are the 8! Stand for? How did you get 2?
OIC
to calculate nCr
use
nCr = n! / [ r! (n-r)! ]
unless you have Pascal's triangle or a binomial coefficient table in front of you
Google binomial coefficient
i! = i * (i-1) * (i-2)..... 1
when you have a division like 19!/8!
you write 10! as 10 * 9 * 8!
so you can cancel the 8!
and be left with 10*9
https://en.wikipedia.org/wiki/Binomial_coefficient
https://en.wikipedia.org/wiki/Factorial
http://en.wikipedia.org/wiki/Binomial_coefficient
as links
http://en.wikipedia.org/wiki/Factorial
You could just turn the terms in the binomial around
(m+3t)^10
then the term we want is just the third term
= C(10,0)m^0 (3t)^10 + C(10,1)m^1 (3t)^9 + C(10,2)m^2 (3t)^8 + ..
C(10,2) = 45 , (Damon explained that)
(3t)^8 = 6561t^8
so we have 45(6561)m^2 t^8
= 295245m^2 t^8
Wolfram confirmation:
look at the third-last term
http://www.wolframalpha.com/input/?i=expand+%283t%2Bm%29%5E10