# How many sets $(A_1,A_2,\cdots, A_k)$ which are subsets of $\{1,2,\cdots ,n\}$

For given natural numbers $n,k$, how many are there $k$-tuples $(A_1,A_2,\cdots ,A_k)$ such that $$A_1\subseteq A_2\subseteq\cdots \subseteq A_k\subseteq \{1,2,3,\cdots ,n\}$$ I've thought to prove by induction on $k$ that the number of $k$-tuples is equal to $$\sum_{t=0}^n{n\choose t}k^t=(k+1)^n$$ Though I have no idea what happens when you add another subset,my idea was when I add another subset to let it be $A_1$ and shift every other subset index by $1$.Then split it into $n+1$ cases such that $|A_2|=t$ for each $t$ from $0$ to $n$.

Maybe there is a better way.

• Which one ???$$A_1\subseteq A_2\subseteq\cdots \subseteq A_k\subseteq \{1,2,3,\cdots ,n\}\\ A_1\subset A_2\subset\cdots \subset A_k\subseteq \{1,2,3,\cdots ,n\}$$? Feb 3 '17 at 18:30
• @Khosrotash The first one Feb 3 '17 at 18:31
• $(k+1)^n$ is the right answer. Hint: assign to each element of $\{1,\dots,n\}$ a number from $0$ to $k$ according to the sequence $A_1\subseteq\dots\subseteq A_k$. Feb 3 '17 at 19:06
• @zhoraster Oh this seems too easy,thanks. Feb 3 '17 at 20:23

The easiest solution is certainly the one in the comments, but we can approach the problem using induction in the following way:

The first observation is that the only important property of the set $\lbrace 1, 2, \dots, n \rbrace$ in the problem is that it has $n$ elements. We would have the same answer if we replaced it with any other set with $n$ elements.

We now use induction on $k$. For the base case, we consider $k = 1$. If we only want a single subset

$$A_1 \subseteq \lbrace 1, 2, \dots, n \rbrace$$

then $A_1$ can be any of the $2^n = {(1+1)}^n$ subsets of $\lbrace 1, 2, \dots, n \rbrace$, and so the number of subsets in this case is indeed ${(k+1)}^n$.

Now suppose that the result is true for some $k$ and for every $n$. We wish to then prove that the number of ways of choosing sets $A_1, A_2, \dots, A_{k+1}$ such that

$$A_1 \subseteq A_2 \subseteq \dots \subseteq A_{k+1} \subseteq \lbrace 1, 2, \dots, n \rbrace$$

is equal to ${(k+2)}^n$.

We count the number of ways of choosing the subsets by considering the number of elements in $A_{k+1}$. Let this number be $m$. Then there are $\binom{n}{m}$ ways to choose the elements in $A_{k+1}$.

Now $A_{k+1}$ is a set with $m$ elements, so by our earlier observation that the set $\lbrace 1, 2, \dots, n \rbrace$ is arbitrary, we can see that once we have chosen $A_{k+1}$, the number of ways of choosing sets $A_1, A_2, \dots, A_k$ such that

$$A_1 \subseteq A_2 \subseteq \dots \subseteq A_k \subseteq A_{k+1}$$

is equal to ${(k+1)}^m$.

Thus the number of ways of choosing sets $A_1, A_2, \dots, A_{k+1}$ such that $$A_1 \subseteq A_2 \subseteq \dots \subseteq A_{k+1} \subseteq \lbrace 1, 2, \dots, n \rbrace$$ and such that $A_{k+1}$ has $m$ elements is equal to $$\binom{n}{m} {(k+1)}^m.$$

We see that the total number of ways of choosing the sets $A_1, A_2, \dots, A_{k+1}$ is then equal to

$$\sum_{m=0}^{n} \binom{n}{m} {(k+1)}^m$$

which by the binomial theorem is equal to ${(k+2)}^n$.

• Thanks for posting your answer (I hope it will be accepted), sparing me the necessity to turn my comment into an answer :) Feb 4 '17 at 19:42