How to prove this; n!/r!(n-r)! +n!/(r-1)!(n-r+1)!=(n-r)!/r!(n-r+1)!
R r267747 New member Joined Oct 1, 2009 Messages 33 Jan 29, 2010 #1 How to prove this; n!/r!(n-r)! +n!/(r-1)!(n-r+1)!=(n-r)!/r!(n-r+1)!
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Jan 29, 2010 #2 Hello, r267747! \(\displaystyle \text{Prove: }\;\frac{n!}{r!\,(n-r)!} +\frac{n!}{(r-1)!\,(n-r+1)!} \;=\; \frac{(n-r)!}{r!\,(n-r+1)!}\) Click to expand... This statement is not true . . . The two factors in the denominator must total the numerator. There must be a typo. \(\displaystyle \text{This statement }is\text{ true: }\;\frac{n!}{r!\,(n-r)!} + \frac{n!}{(r-1)!\,(n-r+1)!} \;=\;\frac{\overbrace{(n+1)!}}{r!\,(n-r+1)!}\) \(\displaystyle \text{Get a common denominator: }\) . . \(\displaystyle \frac{n!}{r!\,(n-r)!}\cdot \frac{n-r+1}{n-r+1} \:+\:\frac{n!}{(r-1)!\,(n-r+1)!}\cdot\frac{r}{r}\) . . . \(\displaystyle =\;\frac{n!(n-r+1)}{r!\,(n-r+1)!} \:+\:\frac{n!(r)}{r!\,(n-r+1)!}\) . . . \(\displaystyle =\;\frac{n!(n-r+1) + n!(r)}{r!(n-r+1)!}\) . . . \(\displaystyle =\;\frac{n!\bigg[(n-r+1) + r\bigg]}{r!\,(n-r+1)!}\) . . . \(\displaystyle =\; \frac{n!(n+1)}{r!\,(n-r+1)!}\) . . . \(\displaystyle =\; \frac{n+1)!}{r!\,(n-r+1)!}\)
Hello, r267747! \(\displaystyle \text{Prove: }\;\frac{n!}{r!\,(n-r)!} +\frac{n!}{(r-1)!\,(n-r+1)!} \;=\; \frac{(n-r)!}{r!\,(n-r+1)!}\) Click to expand... This statement is not true . . . The two factors in the denominator must total the numerator. There must be a typo. \(\displaystyle \text{This statement }is\text{ true: }\;\frac{n!}{r!\,(n-r)!} + \frac{n!}{(r-1)!\,(n-r+1)!} \;=\;\frac{\overbrace{(n+1)!}}{r!\,(n-r+1)!}\) \(\displaystyle \text{Get a common denominator: }\) . . \(\displaystyle \frac{n!}{r!\,(n-r)!}\cdot \frac{n-r+1}{n-r+1} \:+\:\frac{n!}{(r-1)!\,(n-r+1)!}\cdot\frac{r}{r}\) . . . \(\displaystyle =\;\frac{n!(n-r+1)}{r!\,(n-r+1)!} \:+\:\frac{n!(r)}{r!\,(n-r+1)!}\) . . . \(\displaystyle =\;\frac{n!(n-r+1) + n!(r)}{r!(n-r+1)!}\) . . . \(\displaystyle =\;\frac{n!\bigg[(n-r+1) + r\bigg]}{r!\,(n-r+1)!}\) . . . \(\displaystyle =\; \frac{n!(n+1)}{r!\,(n-r+1)!}\) . . . \(\displaystyle =\; \frac{n+1)!}{r!\,(n-r+1)!}\)
