MathLabs

Problem 2

Let m,nm,n be positive integers with n≤mn\le m. Prove that 2nn!≤(m+n)!(m−n)!≤(m2+m)n2^n n!\le\dfrac{(m+n)!}{(m-n)!}\le(m^2+m)^n.
Step 1 of 4: Expand the factorial quotient
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(m+n)!(m−n)!=∏i=1n(m+i)(m−i+1)\frac{(m+n)!}{(m-n)!}=\prod_{i=1}^n(m+i)(m-i+1)
Detailed analysis

The factors from m−n+1m-n+1 through m+nm+n can be paired symmetrically around mm. Therefore (m+n)!(m−n)!=∏i=1n(m+i)(m−i+1)\frac{(m+n)!}{(m-n)!}=\prod_{i=1}^n(m+i)(m-i+1).