How do you evaluate #(7!)/(4!)#? Precalculus The Binomial Theorem The Binomial Theorem 1 Answer Gerardina C. Dec 23, 2016 210 Explanation: Since #n! =n(n-1)(n-2)...3*2*1#, you get #7! =7*6*5*4*3*2*1# and #4! =4*3*2*1# Then #(7!)/(4! )=(7*6*5*cancel(4*3*2*1)) /cancel (4*3*2*1)=210# 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 8293 views around the world You can reuse this answer Creative Commons License