# The intuition behind computing combinations recursively (N choose K)

Note: I have rewritten this article to make it more straightforward, and also I have corrected a lot of grammatical errors.

With this article, I plan to explain the intuition behind the following “recursive” formula. ${N \choose k}={N-1 \choose k-1}+{N-1 \choose k}$
or $C(N,k) = C(N-1,k-1) + C(N-1,k)$

Let’s start by building a mental model, imagine that you have two baskets with you. Let’s call the first as $B_1$, has a capacity of $N$ and is full. The second, $B_2$ has a capacity $k$ (for clarity, $k \leq N$) and empty. For simplicity, let’s assume that all items are distinct and distinguishable.

The idea of choosing $k$ items from $N$ is nothing but counting the number of ways you can pick $k$ items from $B_1$ and put it in $B_2$.

Another critical thing to note is that, while computing combinations, the order of how you choose those $k$ items is unimportant, the only thing that matters is what items you choose. Choosing $k$ items from $N$ is mathematically denoted as ${N \choose k}$ and I shall be using this notation from now on.

One way to approach a problem recursively is to see if it makes sense to decrease the problem size by one and delegate the problem to someone else. Let’s do just that, and pick an item out of $B_1$. Now you can either put it in $B_2$ or keep it aside. Each choice leads to interesting consequences:

Case 1: The item was placed in $B_2$, so now $B_1$ has $N-1$ items in it, and you can only put $k-1$ items in $B_2$. So now you have to choose $k-1$ items from $N-1$ items or in other words ${N-1 \choose k-1}$.

Case 2: The item was placed aside (not in $B_2$), consequently $B_1$ has $N-1$ items in it, but $B_2$ is still empty and you can still put $k$ items in it. This translates to you having to choose $k$ items from $N-1$ items or in other words ${N-1 \choose k}$.

We can conclude that the number of ways of choosing $k$ items from $N$ is the sum of:

• Number of ways of choosing $k-1$ items from $N-1$
• Number of ways of choosing $k$ items from $N-1$.

Thus we have derived the recursive equation, ${N \choose k}={N-1 \choose k-1}+{N-1 \choose k}$.

Are we done? No, not yet. We still haven’t discussed another crucial thing when it comes to recursion, the base cases. Base cases are how the recursive formulae yield an outcome.

Q1) What is ${N \choose N}$?

A: We have only one way of choosing all the items from $B_1$ and putting them into $B_2$. Hence, we can also conclude that ${1 \choose 1}$ is $1$.

Q2) What is ${N \choose 1}$?

A: We have $N$ choices, hence $N$ ways of choosing one item from $N$ items.

Let us make how we stop our recursion a bit smarter by considering the following case:

Q3) What is ${N \choose N-k}$?

A: Choosing $k$ items out of $N$ is the same as not choosing $N-k$ items out of $N$. Hence, you have ${N \choose k} = {N \choose N-k}$. You can derive some interesting results out of this, here are a few

• ${N \choose 0} = {N \choose N} = 1$
• ${N \choose N-1} = {N \choose 1} = N$