Number and Algebra
Functions
Geometry & Trigonometry
Statistics & Probability
Calculus
For x=3x=3x=3, y=2y=2y=2, and k=3k=3k=3, evaluate:
Simplify the factorial expression (x!โy!)!(x!โy!โ1)!โ.\frac{(x!\,y!)!}{\bigl(x!\,y! - 1\bigr)!}\,.(x!y!โ1)!(x!y!)!โ.
Show that (x!โy!+1)!(x!โy!โ1)!=(x!โy!)โ(x!โy!+1)โ.\frac{(x!\,y!+1)!}{\bigl(x!\,y! - 1\bigr)!}=(x!\,y!)\,(x!\,y!+1)\,.(x!y!โ1)!(x!y!+1)!โ=(x!y!)(x!y!+1).
Express (x!โy!)!(x!โy!โk)!\frac{(x!\,y!)!}{\bigl(x!\,y! - k\bigr)!}(x!y!โk)!(x!y!)!โ as a product of kkk consecutive factors. Assume 1โคkโคx!โy!1\le k\le x!\,y!1โคkโคx!y!.
Simplify the expression (x!โy!)!(x!โy!โ1)!โ(xโ1)!โ(yโ1)!โ.\frac{(x!\,y!)!}{\bigl(x!\,y! - 1\bigr)!\,(x-1)!\,(y-1)!}\,. (x!y!โ1)!(xโ1)!(yโ1)!(x!y!)!โ.
Evaluate the following expression.
Evaluate (x!โy!)!(x!โy!โ1)!โ(xโ1)!โ(yโ1)!\frac{(x!\,y!)!}{\bigl(x!\,y! - 1\bigr)!\,(x-1)!\,(y-1)!}(x!y!โ1)!(xโ1)!(yโ1)!(x!y!)!โ for x=2x=2x=2, y=3y=3y=3.
Simplify the expression (x!โy!+2)!(x!โy!)!โ.\frac{(x!\,y!+2)!}{(x!\,y!)!}\,. (x!y!)!(x!y!+2)!โ.
Evaluate (x!โy!)!(x!โy!โ1)!โ(xโ1)!โ(yโ1)!\frac{(x!\,y!)!}{\bigl(x!\,y! - 1\bigr)!\,(x-1)!\,(y-1)!}(x!y!โ1)!(xโ1)!(yโ1)!(x!y!)!โ for x=3x=3x=3, y=4y=4y=4.
Simplify the expression (n+3)!n!โ.\frac{(n+3)!}{n!}\,.n!(n+3)!โ.
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