How to expand (x+y)^5?

where #" "c _k^n= (n!)/(k!(n-k)!)#

and is read "n CHOOSE k equals n factorial divided by k factorial (n-k) factorial".

So #(a+b)^5=a^5+5.a^4.b+10.a^3.b^2+10.a^2.b^3+5.a^1.b^4+b^5#

we notice that the powers of ' a ' keeps decreasing from 5 (which representes ' n ') until it reaches #a^("zero")# at the last term.

also we notice that the power of ' b ' keeps increasing from zero untill it reaches 5 at the last term.

Now the we have to determine the coefficient of each term through the...

#c_k^n= (n!)/(k!(n-k)!)#

first coefficient #c_0^5=(5!)/(0! .5!)=1#

second #c_1^5=(5!)/(1! .4!)=5#

#c_2^5=(5!)/(2! .3!)=10#

#c_3^5=(5!)/(3! .2!)=10#

#c_4^5=(5!)/(4! .1!)=5#

#c_5^5=(5!)/(5!.0!)=1#

but the calculation of combinations can be tedious..so fortunately there is an awesome way to determine the binomial coefficients which is Pascal's triangle