1. Counting sets

To choose an element $a=(a_1,\mathrm{\dots },a_n)\in B_n$ we have $2$ choices for $a_1$, $2$ choices for $a_2$ and so on, making $2\times 2\times \mathrm{\cdots }\times 2=2^n$ choices for the sequence as a whole. ∎

List the elements of $A$ as $x_1,\mathrm{\dots },x_n$. To choose a subset of $A$, we first choose whether to include $x_1$, then choose whether to include $x_2$ and so on. We have two choices for each $x_i$, and thus $2^n$ choices altogether.

Here is another way to say essentially the same thing. Given a binary sequence $a=(a_1,\mathrm{\dots },a_n)$, we define

For example, in the case $n=6$, we have

(In the left hand example, we have ones in positions $1$, $2$ and $3$, so the set is $\{x_1,x_2,x_3\}$. In the right hand example, we have ones in positions $1$, $5$ and $6$, so the set is $\{x_1,x_5,x_6\}$.) This construction gives a one-to-one correspondence between subsets of $A$ and binary sequences, so $|PA|=|B_n|=2^n$. ∎

Suppose we want to choose a sequence $a=(a_1,\mathrm{\dots },a_k)\in F_{k}A$. Then $a_1$ can be any element of $A$, so there are $n$ choices. Then $a_2$ can be any element of $A$ other than $a_1$, so there are $n-1$ choices. Then $a_3$ can be any element other than $a_1$ and $a_2$, so there are $n-2$ choices, and so on. At the last stage, $a_k$ can be any element of $A$ except for $a_1,\mathrm{\dots },a_{k-1}$, so there are $n-(k-1)=n-k+1$ choices. Thus, the overall number of choices is

Note also that

In particular, we have $|F_{n}A|=n!/0!=n!$. In the other hand, if $k>n$, it is clear that we cannot have a list of $k$ distinct elements in $A$, because $A$ has only $n$ elements; so $F_{k}A=\mathrm{\varnothing }$. ∎

If $k>n$ it is clear that $P_{k}A$ is empty so $|P_{k}A|=0=\left(\genfrac{}{}{0pt}{}{n}{k}\right)$. Suppose instead that $0\le k\le n$. Every list $a=(a_1,\mathrm{\dots },a_k)\in F_{k}A$ gives a subset $U_a=\{a_1,\mathrm{\dots },a_k\}\in P_{k}A$, and every subset of size $k$ arises in this way. However, we can reorder the list $a$ in $k!$ different ways, and they all give the same subset. Thus, we have

∎

To specify an element of $B_{nk}$, we just need to specify the $k$ positions in $[1,n]$ where the ones appear. There are $\left(\genfrac{}{}{0pt}{}{n}{k}\right)$ subsets of size $k$ in $[1,n]$, so $|B_{nk}|=\left(\genfrac{}{}{0pt}{}{n}{k}\right)$. ∎

Put $S_n=1+2+\mathrm{\dots }+(n-1)+n$. We can rewrite this with the terms in reverse order as $S_n=n+(n-1)+\mathrm{\cdots }+2+1$. Adding these two equations together, we get

The right hand side consists of $n$ terms, each of which is equal to $n+1$, so the total is $(n+1)n$. It follows that $S_n=(n+1)n/2=\left(\genfrac{}{}{0pt}{}{n+1}{2}\right)$ as claimed. This proof can be illustrated as shown below: there are $S_n$ red dots above the diagonal line and $S_n$ blue dots below it, showing that $2S_n=(n+1)n$.

Alternatively, we can give a proof by induction. For $n=0$ the claim is that $0=\left(\genfrac{}{}{0pt}{}{1}{2}\right)$, which is clear. For $n=1$ the claim is that $1=\left(\genfrac{}{}{0pt}{}{2}{2}\right)$, which is also clear. For $n>1$, we can assume as an induction hypothesis that

Adding $n$ to both sides, we get

as required. ∎

The binomial coefficient $\left(\genfrac{}{}{0pt}{}{n}{k}\right)$ is the number of subsets $A\subseteq [1,n]$ with $|A|=k$. To choose such a subset, we first decide whether we want $n$ to be an element of $A$. If we decide that $n$ should not be an element of $k$, then we just choose $A$ to be a subset of size $k$ in $[1,n-1]$, and there are $\left(\genfrac{}{}{0pt}{}{n-1}{k}\right)$ possibilities for this. If we decide that we do want $n$ to be an element of $A$, then we need to choose a further $k-1$ elements from $[1,n-1]$ to make up the rest of $A$, and there are $\left(\genfrac{}{}{0pt}{}{n-1}{k-1}\right)$ possibilities for this. Thus, we have $\left(\genfrac{}{}{0pt}{}{n-1}{k}\right)+\left(\genfrac{}{}{0pt}{}{n-1}{k-1}\right)$ possibilities for $A$, and this must agree with the number $\left(\genfrac{}{}{0pt}{}{n}{k}\right)$ that we obtained more directly. ∎

Recall that $\left(\genfrac{}{}{0pt}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$. On the top we have

We can also write the $n$ here as $(n-k)+k$, giving $n!=(n-k)\times (n-1)!+k\times (n-1)!$. By substituting this into the definition of $\left(\genfrac{}{}{0pt}{}{n}{k}\right)$, we get

In the first term, we can rewrite $k!$ as $k\times (k-1)!$, and in the second term, we can rewrite $(n-k)!$ as $(n-k)(n-k-1)!$. (These are valid because we are assuming that , so $k,n-k>0$.) This gives

∎

The binomial coefficient $\left(\genfrac{}{}{0pt}{}{n}{k}\right)$ is the number of subsets $A\subseteq [1,n]$ size $k$ in $[1,n]$. To choose such a subset, we can just choose a subset $B\subseteq [1,n]$ of size $n-k$ in $[1,n]$, and take $A=B^c$. This gives a one-to-one correspondence between subsets of size $k$ and subsets of size $n-k$, so $\left(\genfrac{}{}{0pt}{}{n}{k}\right)=\left(\genfrac{}{}{0pt}{}{n}{n-k}\right)$. ∎

∎

We have discussed before that subsets of $[1,n]$ correspond to binary sequences, and it is clear that gappy subsets correspond to gappy sequences, so we will work with binary sequences from now on. We will also assume that $k>0$, leaving the trivial case $k=0$ to the reader. Suppose that we have a gappy sequence $a\in G_{nk}$. The first one in $a$ might appear in the very first position, so it need not be preceded by a zero. However, there are $k-1$ more ones, and by the gappy condition, each of them must have a zero immediately before it. The ones and these adjacent zeros take $2k-1$ slots altogether, so we must have $n\ge 2k-1$. This shows that $G_{nk}=\mathrm{\varnothing }$ if ; we will assume that $n\ge 2k-1$ from now on. If we delete these zeros, we get a binary sequence of length $n-k+1$ containing $k$ ones, or in other words, and element of $B_{n-k+1,k}$. On the other hand, if we are given an element of $B_{n-k+1,k}$, then we can get an element of $G_{nk}$ by inserting zeros to the left of all the ones, except for the first one. Thus, we have a one-to-one correspondence between $G_{nk}$ and $B_{n-k+1,k}$, showing that $|G_{nk}|=|B_{n-k+1,k}|=\left(\genfrac{}{}{0pt}{}{n-k+1}{k}\right)$.

Here are some examples of elements of $G_{73}$, and the corresponding elements of $B_{53}$:

∎