# Show that $\frac{x(x-1) \dots (x-n+1)}{n!} \in \mathbb{Z}$ with $x \in \mathbb{Z}$ [duplicate]

Problem: Let polynomial $$Q_n (x) = \frac{x(x-1) \dots (x-n+1)}{n!} \in R[x]$$ for some ring $$R$$. Show that $$\forall t \in \mathbb{Z}, Q_n (t) \in \mathbb{Z}$$.

My solution: For each $$t \in \mathbb{Z}$$, we have $$t-1, t-2, \dots, t-n+1 \in \mathbb{Z}$$, so $$\frac{t(t-1) \dots (t-n+1)}{n!} \in \mathbb{Z}$$ and $$Q_n (t) \in \mathbb{Z}$$.

Please check my solution. Thank all!

## marked as duplicate by Servaes, darij grinberg, metamorphy, Yanior Weg, mrtaurhoMay 22 at 20:39

• How do you know that each of the factors in the numerator is a multiple of one of the numbers from $1$ through $n$? – Aniruddha Deshmukh May 22 at 13:01
• You have just stated that all factors in the numerator are integers. But instead, you should show that the numerator is a multiple of $n!$. To do this, you can read the answers here: The product of $n$ consecutive integers is divisible by $n$ factorial. – Minus One-Twelfth May 22 at 13:02
• You proved nothing. The product of $n$ integers needn't be a multiple of $n!$. – Yves Daoust May 22 at 13:02
• The combinatorial argument: $Q_n(x)=\binom xn$ is the number of ways of choosing $n$ out of $x\ge n$ objects. For $0\le x<n$, one of the factors of the numerator is $0$. For $x<0$, the argument is similar to the one for $x\ge n$. – Shubham Johri May 22 at 13:03
• I knew this formulate $\binom{m}{n} = \frac{n(n-1)\dots(n-m+1)}{m!}$. But with this formulate, the prove is clear. – Minh May 22 at 13:09

where $$(\cdot )_{k}$$ is the Pochhammer symbol, here standing for a falling factorial. So,
$$Q_n (x) = \frac{x(x-1) \dots (x-n+1)}{n!} = \frac {(x)_n} {n!} = \frac {\frac {x!} {(x-n)!}} {n!} = \frac {x!} {n! (x-n)!} = \binom x n \in R[x]$$ Now we know that $$x$$ is an integer so: $$\forall t \in \mathbb{Z}, Q_n (t) \in \mathbb{Z}$$.