Dividing factorials I'm told that <mstyle displaystyle="true" scriptlevel="0">

Kolten Conrad 2022-06-30 Answered
Dividing factorials
I'm told that $\frac{\left(n+1\right)!}{\left(n+2\right)!}$ simplifies to $\frac{1}{n+2}$, but I dont understand how this works.
Could someone explain the theory of how to divide factorials like this?
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Answers (2)

Sophia Mcdowell
Answered 2022-07-01 Author has 14 answers
HINT:
Example, set $n=5$
$\frac{\left(5+1\right)!}{\left(5+2\right)!}=\frac{6!}{7!}=\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{1}{7}\cdot \frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{1}{7}\cdot 1=\frac{1}{7}=\frac{1}{5+2}$
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Cristopher Knox
Answered 2022-07-02 Author has 6 answers
For an algebraic proof:
$\frac{\left(n+1\right)!}{\left(n+2\right)!}=\frac{\left(n+1\right)n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\dots 5\cdot 4\cdot 3\cdot 2\cdot 1}{\left(n+2\right)\left(n+1\right)n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\dots 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{\overline{)\left(n+1\right)n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\dots 5\cdot 4\cdot 3\cdot 2\cdot 1}}{\left(n+2\right)\overline{)\left(n+1\right)n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\dots 5\cdot 4\cdot 3\cdot 2\cdot 1}}=\frac{1}{\left(n+2\right)}$
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