Why does $\binom {3} {1} =\binom {3} {2}$?

Why does $$\binom {3} {1} =\binom {3} {2}$$? Is it just a coincidence?

Always $$\binom{n}{k}=\binom{n}{n-k}$$ (when $$0 \le k \le n.)$$ Because choosing a $$k$$ element set from $$n$$ things is the same as choosing its complement.

So also $$\binom{10}{3}=\binom{10}{7}$$ and one can generate lots of these.

The binomial coefficient is calculated as :

$$\binom{n}{k}= \frac{n!}{k!(n-k)!}$$

It's easy to see why the two binomial coefficients mentioned are equal, based on the definition.

In general, it always holds that :

$$\binom{n}{k}=\binom{n}{n-k}, \quad 0\leq k \leq n$$

Essentialy, what this means is that choosing an element $$k$$ from $$n$$ elements is the same as not choosing $$n-k$$ elements from $$n$$ elements.

Because to choose $$1$$ from $$3$$ things it's the same not to choose $$2$$ from $$3$$ things.

No, it’s not. We have

$${n \choose r} = {n \choose n-r}$$

for non-negative values of $$r$$ with $$r \leq n$$, which can be verified through the definition $${n \choose r} = \frac{n!}{r!(n-r)!}$$:

$$\frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!(n-(n-r))!}$$

$$\frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!r!}$$

$$\binom{3}{1}$$ is calculating the number of ways to choose one object out of a set of three. $$\binom{3}{2}$$ is calculating the number of ways to choose two objects out of a set of three.

Imagine I'm looking at a group of three people: Alice, Bob, and Charlotte. Let's say I want to choose two of them, and I've decided that those two are going to be Alice and Bob. There are two ways for me to say that - one way is the obvious, "I'm going to choose Alice and Bob." The other way is to talk about who I'm not choosing: "I'm going to choose everyone except Charlotte."

More generally, when picking two objects out of a group of three, we can phrase that as deciding which one is going to be left over. In other words, choosing which two to take is the same as choosing which one not to take. There are $$\binom{3}{2}$$ ways to choose which two to take; there are $$\binom{3}{1}$$ ways to choose which one not to take. Since those mean the same thing, $$\binom{3}{1} = \binom{3}{2}$$.

It is not just a coincidence. Nothing I just said really had anything to do with the numbers $$3$$, $$2$$, and $$1$$. What mattered was that when $$2$$ were chosen out of a group of $$3$$, $$1$$ was left over. We can generalize that: when $$k$$ are chosen out of a group of $$n$$, $$n - k$$ will be left over; so choosing $$k$$ out of $$n$$ is the same as excluding $$n - k$$ out of $$n$$. In other words:

$$\binom{n}{k} = \binom{n}{n - k}$$

Imagine you have a group of $$n$$ objects, and you choose $$k$$ of those objects to remove.

However, instead of considering it like this, consider it as dividing the group of $$n$$ objects into two groups, one of which has $$k$$ objects (and consequently, the other has $$(n-k)$$ objects).

We can flip this scenario and remove $$(n-k)$$ objects, leaving $$k$$ behind.

Can you see how these two scenarios are the same?

Short Answer: No, These numbers are related to the Pascal's triangle which is symmetrical. The third row is 1 , 3 , 3, 1. This states that (3 Choose 1) = (3 Chose 2): which you stated, also: (3 Choose 0) = (3 Choose 3). You can continue this for any number of items.

• I think it would be a bit clearer to use a complete sentence: "Short Answer: No, it is not a coincidence." Then follow with the explanation as above. "No" is doubtless clear to you in this sense, but to other Readers "no" might mean an objection to some other aspect of the Question posed. – hardmath Feb 25 '19 at 4:00