First post in Stack Exchange and feel bad to be in need of help. But, I'm having a hard time understanding this one or rather showing the argument.
$\binom{n}{k} = \binom{n-2}{k-2} + 2\binom{n-2}{k-1} + \binom{n-2}{k}$
So, the argument that I'm thinking of: How many ways to form a committee of $k$ members from a set of $n$ people.
Answer 1: Select $k$ members to form a committee from $n$ people; therefore there are $\binom{n}{k}$ ways.
Answer 2: (This is where I'm stuck. The left side is confusing me)
3 distinct cases to consider. 1) Two members from the committee left so $n-2$ of people left. 2) One member from the committee left and two...(idk what I'm doing here). 3) Two people left from the set of $n$, but it didn't affect the $k$ members of the committee.
I'm not sure if I'm somewhat near or if I'm explaining the RHS well.
We have a group of $n\ge 2$ people, including Alicia and Beti. We want to form a committee of $k\ge 2$ people. Of course there are $\binom{n}{k}$ ways to choose the committee.
Let us count another way. There are $\binom{n-2}{k-2}$ (this is in fact $\binom{n-2}{k-2}\binom{2}{2}$) committees of $k$ people that have both Alicia and Beti.
There are $\binom{n-2}{k-1}$ committees with A but not B, and the same number with B and not A, for a total of $2\binom{n-2}{k-1}$ (this is in fact $\binom{n-2}{k-1}\binom{2}{1}$).
And finally there are $\binom{n-2}{k}$ with neither A nor B (this is in fact $\binom{n-2}{k}\binom{2}{0}$).
Remark: Maybe too bureaucratic! We have $n$ doughnuts, of which $1$ is Almond, $1$ is Berry, and the other $n-2$ are plain. We want to choose $k$ doughnuts to eat right now. The rest of the story is basically the same as the committee story.