Proof that a Combination is an integer

Question

From its definition a combination $\binom{n}{k}$, is the number of distinct subsets of size $k$ from a set of $n$ elements.

This is clearly an integer, however I was curious as to why the expression

$$\frac{n!}{k!(n-k)!}$$ always evaluates to an integer.

So far I figured:

$n!$, is clearly divisible by $k!$, and $(n-k)!$, individually, but I could not seem to make the jump to proof that that $n!$ is divisible by their product.

Answer 1

Well, one noncombinatorial way is to induct on $n$ using Pascal's triangle; that is, using the fact that ${n \choose k} = {n-1 \choose k - 1} + {n-1 \choose k}$ (easy to verify directly) and that each ${n - 1 \choose 0}$ is just $1$.

Answer 2

See my post here for a simple purely arithmetical proof that every binomial coefficient is an integer. The proof shows how to rewrite any binomial coefficient fraction as a product of fractions whose denominators are all coprime to any given prime $\rm\:p.\,$ This implies that no primes divide the denominator (when written in lowest terms), therefore the fraction is an integer.

The key property that lies at the heart of this proof is that, among all products of $\rm\, n\,$ consecutive integers, $\rm\ n!\ $ has the least possible power of $\rm\,p\,$ dividing it - for every prime $\rm\,p.\,$ Thus $\rm\ n!\ $ divides every product of $\rm\:n\:$ consecutive integers, since it has a smaller power of every prime divisor. Therefore $$\rm\displaystyle\quad\quad {m \choose n}\ =\ \frac{m!/(m-n)!}{n!}\ =\ \frac{m\:(m-1)\:\cdots\:(m-n+1)}{\!\!n\:(n-1)\ \cdots\:\phantom{m-n}1\phantom{+1}}\ \in\ \mathbb Z$$

Answer 3

As Jonas mentioned, it counts something so it has to be a natural number.

Another way is to notice that product of $m$ consecutive natural numbers is divisible by $m!$.(Prove this!)

So if we write $n! = n(n-1)(n-2) \cdots (k+1) \times (k!)$, we find that $k!$ divides $k!$ and

$n(n-1)(n-2) \cdots (k+1)$ is a product of $(n-k)$ consecutive natural numbers and hence $(n-k)!$ divides it.

Answer 4

According to this, the highest power of a prime number, $p, $ that divides $N!$ is $$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$

Since $\dbinom{N}{M} = \dfrac{N!}{M! (N-M)!}\,, $ the question then becomes, is $s_p(M!) + s_p((N-M)!) \le s_p(N!)$?

Clearly $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor \le x + y. $ Since $\left \lfloor x+y \right \rfloor$ is the greatest integer less than or equal to $x + y, $ it follows that $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor \le \left \lfloor x+y \right \rfloor,$

Hence $$\left \lfloor \dfrac{M}{p^{\,i}} \right \rfloor + \left \lfloor \dfrac{N-M}{p^{\,i}} \right \rfloor \le \left \lfloor \dfrac{N}{p^{\,i}} \right \rfloor$$

for all $i$. Summing, we get $s_p(M!) + s_p((N-M)!) \le s_p(N!)$

It follows that $\dbinom{N}{M}$ is an integer.

Answer 5

Zarrax's answer is a good start, but to show that $\frac{n!}{k!\,(n-k)!}\in\mathbb{Z}$ we should really show that the definition by Pascal's Rule and $\frac{n!}{k!\,(n-k)!}$ agree.

First, the edge cases match, $\binom{n}{0}=\binom{n}{n}=1$: $$ \frac{n!}{0!\,n!}=1\qquad\text{and}\qquad\frac{n!}{n!\,0!}=1 $$

Next, Pascal's Rule fits, $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$: $$ \begin{align} \frac{(n-1)!}{(n-k-1)!\,k!}+\frac{(n-1)!}{(n-k)!\,(k-1)!} &=\frac{(n-k)\,(n-1)!}{(n-k)!\,k!}+\frac{k\,(n-1)!}{(n-k)!\,k!}\\ &=\frac{n\,(n-1)!}{(n-k)!\,k!}\\ &=\frac{n!}{(n-k)!\,k!} \end{align} $$ Therefore, by induction on $n$, $$ \binom{n}{k}=\frac{n!}{k!\,(n-k)!} $$

Answer 6

Another way to think of it is combinatorially, which is of course the motivation.

Let $1\le k\le n$. Consider the set $S$ of all sequences of $k$ distinct numbers among $\{1,\dots,n\}$. The size of $S$ is $n\cdot (n-1)\cdots (n-k+1)=\frac{n!}{(n-k)!}$. Say two sequences in $S$ are equivalent if the underlying set of elements is the same. Then each equivalence class has exactly $k!$ elements since any choice of $k$ elements admits $k!$ permutations. So the set $S$ can be partitioned into sets each of which has size $k!$. So $|S|$ is divisible by $k!$.

Answer 7

Consider the set

$S=\Big\{k\in\mathbb{N}:(\forall m,n\in \mathbb{N})\left(0\leq m\leq n\wedge m+n=k\Rightarrow \binom{n}{m}\in\mathbb{N}\right)\Big\}$

1) If $0\leq m\leq n$ and $m+n=0$, then $m=n=0\Rightarrow \binom{n}{m}=\binom{0}{0}=1\in \mathbb{N}\Rightarrow 0\in S$

2) If $0\leq m\leq n$ and $m+n=1$, then $m=0\wedge n=1\Rightarrow \binom{n}{m}=\binom{1}{0}=1\in\mathbb{N}\Rightarrow 1\in S$

Assume that $p\in S,\, \forall p\leq k$, then $0\leq m\leq n\wedge m+n=p\leq k\Rightarrow \binom{n}{m}\in\mathbb{N}$

If $m=0$, then $\binom{n}{m}=\binom{n}{0}=1\in \mathbb{N},\, \forall n\in\mathbb{N}$

If $m=n$, then $\binom{n}{m}=\binom{n}{n}=1\in \mathbb{N},\, \forall n\in\mathbb{N}$

Therefore we can only consider the cases in which $0<m<n$.

If $0<m<n$ and $m+n=k+1$, then $m+(n-1)=k$ which implies, by induction hypothesis, that $\binom{n-1}{m}\in \mathbb{N}$ since $0<m< n\Rightarrow 0<m\leq n-1$.

$(m-1)+(n-1)=k-1\leq k\Rightarrow \binom{n-1}{m-1}\in\mathbb{N}$ since $0<m<n\Rightarrow 0\leq m-1<n-1 $(also by induction hypothesis)

By Pascal's rule we have

$\binom{n}{m}=\binom{n-1}{m}+\binom{n-1}{m-1}$

We saw that $\binom{n-1}{m}$ and $\binom{n-1}{m-1}$ are natural number, therefore $\binom{n}{m}$ is also a natural number, by pascal's rule.

Therefore $k+1\in S$ which implies, by strong induction, that $S=\mathbb{N}$. Hence $\binom{n}{m}\in\mathbb{N},\, \forall 0\leq m\leq n$.

Answer 8

I will just fill in the details to the proof suggested by Zarrax, above. (In other words, I'll prove the statement by induction, using Pascal's rule.)

In what follows, $\mathbb{N}_{0}$ is the set of non-negative integers, $\mathbb{N}_{0}=\left\{0,\,1,\,2,\,\ldots\right\}$.

Statement: if $n,\,k\in \mathbb{N}_{0}$ with $k\leqslant n$, then $n\choose{k}$ is an integer.

Proof. First, a direct computation shows for all $n\in\mathbb{N}_{0}$, we have ${n\choose{0}}={n\choose{n}}=1$. After all, ${n\choose{0}}=\frac{n!}{0!\,(n-0)!}=\frac{n!}{1\times\,n!}=1$, and similarly for ${n\choose{n}}$.

This already shows that the statement is true for $n=0$ and $1$, since for such $n$, we have that $k$ can only be $0$ or $n$. We now have to prove that the statement is also true for $n>1$.

To proceed, we will need Pascal's rule in the following form:

Let $a,\,k\in\mathbb{N}_{0}$ such that $a\geqslant 2$ and $1\leqslant k \leqslant a-1$. Then ${a\choose{k}}={a-1\choose{k-1}}+{a-1\choose{k}}$.

(This is a weaker form than the one stated e.g. here because of the added restrictions $a\geqslant 2$ and $k \leqslant a-1$.) Note that in Pascal's rule as we've stated it, $k-1$ and $a-1$ are never less than 0, while both $a-k$ and $k$ are strictly greater than zero. Thanks to that, this version of Pascal's rule can be proven by a straightforward computation: ${a-1\choose{k-1}}+{a-1\choose{k}}=\frac{(a-1)!}{(k-1)!(a-k)!}+\frac{(a-1)!}{k!(a-k-1)!}=\frac{(a-1)!}{(k-1)!(a-k-1)!}\left(\frac{1}{a-k}+\frac{1}{k}\right)=\frac{(a-1)!}{(k-1)!(a-k-1)!} \frac{k+(a-k)}{(a-k)k}=\frac{(a-1)!}{(k-1)!(a-k-1)!} \frac{a}{(a-k)k}={a\choose{k}}\,.$

Note that because of the restrictions on $a$ and $k$, every term that appears in the computation is well-defined.

Finally, let us show by induction that the statement is true for all $n\in\mathbb{N}_{0}$. As we said, the cases $n=0$ and $1$ have already been proven. We can take $n=1$ as the base case for induction.

Now we assume that the statement holds for $n=m$, and prove that it then must also hold for $n=m+1$.

To do that, we must prove that $m+1\choose{k}$ is an integer for all $0\leqslant k \leqslant m+1$. The cases $k=0$ and $k=m+1$ follow by the direct computation we've done above, so we only have to prove the statement for all $k$ such that $1\leqslant k \leqslant m$. But in that case, Pascal's rule (as we've stated and proven it) applies for $a=m+1$ (note that since $n=1$ is our base case, $m+1$ is at least $2$; also, $a-1=m$), and we can write ${m+1\choose{k}}={m\choose{k-1}}+{m\choose{k}}$. But both ${m\choose{k-1}}$ and ${m\choose{k}}$ are integers by the induction hypothesis; so ${m+1\choose{k}}$ is a sum of two integers, thus an integer. $\square$

Answer 9

One way to get there: any product of $k$ consecutive positive integers $d := \frac{n!}{(n-k)!} = \prod^k (n-i+1)$ is divisible by $k!$.

You can assure yourself of this by considering divisibility by the primes and prime powers up to $k$. For example, $k!$ has $\lfloor k/p\rfloor$ multiples of $p$, and $d$ will have either the same number of multiples of $p$ or one more, since the "first" multiple of $p$ in $k!$ will be at $p$, whereas it may be earlier in $d$.

The quantity I have denoted by $d$ is the $k$-descending factorial of $n$, sometimes written $n^{\underline k}$ or $(n)_k$.

Answer 10

The proofs I see here so far seemed rather complicated to me, so I came up with what seemed like a simpler proof. However, it proved to have gaps in reasoning; so I came up with an even simpler proof:

Theorem: The product of any k consecutive positive integers [k>=1] is always divisible by k!.

Proof:

1. (m+1)...(m+k) = (m+k)!/m! (By definition of "factorial"; note that left side, right side, numerator, and denominator are all integers.)
2. (m+1)...(m+k) = ((m+k)!/((k!)(m!)))k! (Multiplied top and bottom of fraction by integer k!; therefore left side, right side, numerator, denominator, fraction, and k! are all integers.)
3. Hence, (m+1)...(m+k) is always an integer multiple of k!
4. Hence, (m+1)...(m+k) is always divisible by k!, QED.

Corollary: n!/((n-k)!k!) [n>=1,k>=1,k<=n] is always an integer.

Proof:

1. n!/((n-k)!k!) = ((n-k+1)(n-k+2)...(n-k+k)) / k!
2. This is a product of k consecutive integers, divided by k!.
3. Any product of k consecutive integers is always divisible by k!, due to the theorem I just proved above.
4. Therefore n!/((n-k)!k!) is an integer, QED.

[Note: the theorem and its corollary are also valid for n=0, k=0, m=0 if we stipulate certain definitions:

1. "The product of 0 consecutive integers" = 1
2. 0 is "divisible" by every positive integer
3. "0!" = 1.

Without those stipulations, the theorem & corollary fails if k, m, or n is 0.]