Sum of binomial coefficients $\sum _{ x=r-2 }^{ n-2 } \binom{x}{r-2}$

Question

$$\sum _{ x=r-2 }^{ n-2 } \binom{x}{r-2}$$ I can't find the sum of the following series. I would appreciate if anyone can show me this problem's solution.

Answer 1

You want to count the number of binary sequences which have a quantity of digits ranging from $r-2$ to $n-2$ which have exactly $r-2$ times the digit $n-2$. These sequences are in bijective correspondance with the sequences with $n-1$ digits that contain the digit $1$ exactly $r-1$ times. The bijection is as follows: Given a sequence of length $n-1$ with exactly $r-1$ apparitions of the digit $1$ We do the following: Locate the rightmost digit $1$. Delete it and all the digits to the right of it.

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In general we have the following:

$\sum\limits_{k=a}^b\binom{k}{a}=\sum\limits_{k=0}^b\binom{k}{a}=\binom{b+1}{a+1}$

Answer 2

The terms of a sum lie on a diagonal in Pascal's triangle stretching from $\binom{r-2}{r-2}$ and southwest towards $\binom{n-2}{r-2}$. The definition of Pascal's triangle directly implies that it contains the sum of each such truncated diagonals, located at one place southeast of its bottom end -- that is, in this case, at $\binom{n-1}{r-1}$.

To prove this, try induction on $n-r$.

Combinatorially: The number of ways to line up between $r-2$ and $n-2$ boxes with a ball in $r-2$ of them is the same as lining up $n-1$ boxes, put balls in $r-1$ of them, and then remove the box with the leftmost ball, and all boxes to the left of it.