Pascal's rule : I don't understand why we can do this...

Question

So I understand that they are equal, but I can't get my head around the wording. Why can we choose "1 special element, and $k − 1$ from the remaining $n − 1$, or choose all $k$ from the $n − 1$"?

To choose $k$ elements from $n$ we can either choose 1 special element, and $k-1$ from the remaining $n-1$, or choose all $k$ from the $n-1$. This [sic] we have Pascal's Rule $$C_k^n = C_{k-1}^{n-1} + C_k^{n-1}. $$ This leads on to the so-called Pascal's Triangle.

Sorry if this basic but sometimes I have an issue getting around wording I'm hope there is a simpler wording for it!

Answer 1

You want to pick 5 children from a group of 9 to form a team. You can either pick Billy and then pick 4 out of the remaining 8, or not pick Billy and pick all 5 children out of the remaining 8. Thus ${9 \choose 5} = {8 \choose 4} + {8 \choose 5}$. The general argument is the same.

Answer 2

We can also see the role of the special element and the kind of selection by a purely algebraic approach. We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a polynomial. We can write this way \begin{align*} \binom{n}{k}=[x^k](1+x)^n \end{align*}

We obtain \begin{align*} \binom{n}{k}&=[x^k](1+x)^n\\ &=[x^k](1+\color{blue}{x})(1+x)^{n-1}\\ &=[x^k](1+x)^{n-1}+[x^k]\color{blue}{x}(1+x)^{n-1}\\ &=[x^k](1+x)^{n-1}+[x^{k-1}](1+x)^{n-1}\\ &=\binom{n-1}{k}+\binom{n-1}{k-1} \end{align*} The special element $\color{blue}{x}$ is marked blue.

Answer 3

To see the correspondence, note that the number $C_k^n = \binom{n}{k}$ is the number of a teams we can construct of size $k$ with $n$ players. Now assign one of the players as "special". This player can be either included or not included in the team, so if it's included there are $C_{k-1}^{n-1} = \binom{n-1}{k-1}$ ways to choose the rest players. If he's not we can choose the team in $C_k^{n-1} = \binom{n-1}{k}$ ways.

The punchline is that the special player must be included OR not included in the team and they are two disjoint cases, so we can find the number of combinations in both cases and add them to obtain the total sum.

Answer 4

It just explains that, once you've marked one (fixed) element, choosing $k$ elements from a set of $n$ can be done in two (incompatible) ways:

• either this marked element is in your choice, and there are $\dbinom{n-1}{k-1}$ ways to choose the $k-1$ other among the remaining $n-1$;
• or it is not, and you have $\dbinom{n-1}k$ possible ways.

Answer 5

We want to make a bundle of $k$ books (order does not matter) from a box of $n$ books. By definition there are $C^n_k$ ways to do this.

Now imagine all $C^n_k$ possible bundles of $k$ books you could have obtained, and split them into two collections: one collection has all possible bundles of $k$ books that contain "A Tale of Two Cities," and the other collection has all possible bundles of $k$ books that do not contain "A Tale of Two Cities." You can reason that the first collection has $C^{n-1}_{k-1}$ sets, and the second has $C^{n-1}_k$ sets. Together they must add to $C^n_k$ total bundles.

Answer 6

You might interpret this as follows: you start with $n$ objects, one of which is special (perhaps you have $n$ Sneeches, one of which is star-bellied). You want to choose $k$ of those Sneeches for... something. There are two ways you can do this:

1. You could get a collection of $k$ Sneeches which includes the star-bellied Sneech. Once you put that Sneech into your collection, there are $n-1$ Sneeches remaining, of which you want to choose $k-1$. This gives the term $C_{k-1}^{n-1}$

2. You could get a collection that does not include the star-bellied Sneech. To do this, you must choose $k$ from among the $n-1$ Sneeches that do not have stars on their bellies. This gives the $C_{k}^{n-1}$ term.

As these events are disjoint, this gives a total of $$ C_{k-1}^{n-1} + C_{k}^{n-1} $$ ways of choosing $k$ Sneeches from a group of $n$, one of which is star-bellied. But we could equivalently just ask how many ways there are of choosing $k$ of the $n$ Sneeches, disregarding the fact that one has a star on its belly. There are $C_{k}^{n}$ ways of doing this, which gives the desired identity.

Answer 7

Choosing $k$ from $n$ gives the same number of possibilities as choosing just $k-1$ from just $n-1$ plus the possibilities of preselecting one $k-1$ out of $n-1$ situation, and these are $k$ from $n-1$.

Answer 8

Color one the elements, say with red. Now among chosen elements there can be red one or not. Now how many ways are there to choose them if red is among them? The answer is $n-1 \choose k-1$. And on how many ways red is not among chosen elements? $n-1 \choose k$, and thus the formula.