How do you evaluate #(12!)/(8!4!)#? Precalculus The Binomial Theorem The Binomial Theorem 1 Answer Somebody N. Apr 18, 2018 #color(blue)(495)# Explanation: #(12!)/(8!4!)=(12xx11xx10xx9xx8!)/(8!xx4xx3xx2xx1....)# #=(12xx11xx10xx9xx8!)/(8!xx12xx2xx1....)# #=(cancel(12)xx11xx10xx9xxcancel(8!))/(cancel(8!)xxcancel(12)xx2xx1....)=(11xx10xx9)/(2)=990/2=495# Answer link Related questions What is the binomial theorem? How do I use the binomial theorem to expand #(d-4b)^3#? How do I use the the binomial theorem to expand #(t + w)^4#? How do I use the the binomial theorem to expand #(v - u)^6#? How do I use the binomial theorem to find the constant term? How do you find the coefficient of x^5 in the expansion of (2x+3)(x+1)^8? How do you find the coefficient of x^6 in the expansion of #(2x+3)^10#? How do you use the binomial series to expand #f(x)=1/(sqrt(1+x^2))#? How do you use the binomial series to expand #1 / (1+x)^4#? How do you use the binomial series to expand #f(x)=(1+x)^(1/3 )#? See all questions in The Binomial Theorem Impact of this question 3355 views around the world You can reuse this answer Creative Commons License