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 4 of 4: Multiply the bounds
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∏i=1n2i=2nn!≤∏i=1n(m+i)(m−i+1)≤∏i=1n(m2+m)=(m2+m)n\prod_{i=1}^n2i=2^n n!\le\prod_{i=1}^n(m+i)(m-i+1)\le\prod_{i=1}^n(m^2+m)=(m^2+m)^n
Detailed analysis

All paired factors are positive, so multiplying the termwise inequalities preserves the order. Using the product identity from the first step gives exactly 2nn!≤(m+n)!(m−n)!≤(m2+m)n2^n n!\le\frac{(m+n)!}{(m-n)!}\le(m^2+m)^n.