How do use the binomial theorem to calculate #""^8C_5#?

#color(white)(a)^nC_r=(n!)/(r!(n-r)!#

so

#color(white)(A)^8C_5=(8!)/(5!(8-5)!)#

#=(8xx7xx6xxcancel(5!))/(cancel(5!)3!)#

#=(8xx7xxcancel(6))/(cancel(3!))#

#=56#

#"using the definition of "^nC_r#

#•color(white)(x)^nC_r=(n!)/(r!(n-r)!)#

#"where "n! =n(n-1)(n-2)(n-3)......xx3xx2xx1#

#rArr^8C_5#

#=(8!)/(5!3!#

#=(8xx7xx6xxcancel(5xx4xx3))/(cancel(5!)xx3xx2xx1#

#=(8xx7xx6)/6=56#

This seems a curious question to me, since normally you would use #""^8C_5 = (8!)/(3!5!)# to find a binomial coefficient, but let's give it a try.

Applying the Binomial theorem to #f(x) = (x+1)^8# we have:

#(x+1)^8 = ""^8C_0 x^8 + ""^8C_1 x^7 + ""^8C_2 x^6 + ""^8C_3 x^5 + ""^8C_4 x^4 + ""^8C_5 x^3 + ""^8C_6 x^2 + ""^8C_7 x + ""^8C_8#

Taking the derivative #3# times, we find:

#8 * 7 * 6 (x+1)^5#

#= f^((3)) (x)#

#= 8 * 7 * 6 * ""^8C_0 x^5 + 7 * 6 * 5 * ""^8C_1 x^4 + 6 * 5 * 4 * ""^8C_2 x^3 + 5 * 4 * 3 * ""^8C_3 x^2 + 4 * 3 * 2 * ""^8C_4 x + 3 * 2 * 1 * ""^8C_5#

So:

#8 * 7 * 6#

#= 8 * 7 * 6 * ((color(blue)(0))+1)^5#

#= f^((3)) (0)#

#= 8 * 7 * 6 * ""^8C_0 (color(blue)(0))^5 + 7 * 6 * 5 * ""^8C_1 (color(blue)(0))^4 + 6 * 5 * 4 * ""^8C_2 (color(blue)(0))^3 + 5 * 4 * 3 * ""^8C_3 (color(blue)(0))^2 + 4 * 3 * 2 * ""^8C_4 (color(blue)(0)) + 3 * 2 * 1 * ""^8C_5#

#= 3 * 2 * 1 * ""^8C_5#

So:

#""^8C_5 = (8 * 7 * 6) / (3 * 2 * 1) = 56#

Alternatively, you might recognise that the method of constructing Pascal's triangle relates precisely to the values of the coefficients of #(a+b)^n# and hence deduce that the appropriate entry in Pascal's triangle must be the binomial coefficient.

The row of Pascal's triangle starting #1, 8,...# contains the coefficients for #(a+b)^8#, with the coefficient of #a^3b^5# being #56#, a.k.a. #""^8C_5# ...