Binomial coefficient proof Prove that for any $0\lt r\lt n$ we have $$\binom nr=\binom{n-1}{r-1}+\binom{n-1}r.$$
How do prove this and what step do i take in order for it to be true? 
 A: Ooh, one of those binomial coefficient questions that can be answered by giving a small story. I like those.
The concept to think about in this case is that of a bowl filled with $n$ balls and one of those $n$ balls is coloured pink, while the rest ($n-1$ in total) is just plain, boring white. If you take $r$ balls from this bowl, either the pink ball is one of those (in $\binom{n-1}{r-1}$ cases; can you see why?) or it isn't (in $\binom{n-1}{r}$ cases).
A: This is simple. $(_r^n)$ corresponds to the number in the $(r+1)$th position of the $(n+1)$th row of pascal's triangle. Pascal's triangle is generated by, for each position, adding the numbers in the positions above it, which happen to be $(_{r-1}^{n-1})$ and $(_r^{n-1})$. 
A: By definition, the inomial coefficient  $n\choose k$ is the coeffient of $x^k$ in the expansion of the binomial $(1+x)^n$.
From $$(1+x)^n=(1+x)^{n-1}\cdot (1+x)=(1+x)^{n-1} + x\cdot(1+x)^{n-1}$$
(for $n\ge 1$) we see that the coefficient of $x^r$ in $(1+x)^r$ is the sum of the coefficients of $x^{r}$ and of $x^{r-1}$ in $(1+x)^{n-1}$, in other words
$$ {n\choose r}={n-1\choose r}+{n-1\choose r-1}.$$
A: Suppose you have a group of $n-1$ boys. You want to select $r-1$ or $r$ boys out of them.
You are too lazy to write $$\binom{n-1}{r-1}+\binom{n-1}{r}$$
So, you take Justin Bieber and put him with the $n-1$ boys and then select $r$ people. 
$$\binom{n}{r}$$
If Justin Bieber comes up in your list, you can claim that he is not a guy you selected $r-1$ boys. If he doesn't, well, you selected $r$ boys.
A: Apply the definition to $\binom{n-1}{r}+\binom{n-1}{r-1}$:
\begin{align*}
\frac{(n-1)!}{r!(n-1-r)!}+\frac{(n-1)!}{(r-1)!(n-1-(r-1))!}&=\\[1.5ex]
\frac{(n-1)!}{r!(n-1-r)!} \cdot \frac{n-r}{n-r}+\frac{(n-1)!}{(r-1)!(n-r)!} \cdot \frac{r}{r}&=\\[1.5ex]
\frac{(n-r)(n-1)!}{r!(n-r)!}+\frac{r(n-1)!}{r!(n-r)!}&=\\[1.5ex]
\frac{(n-r)(n-1)!+r(n-1)!}{r!(n-r)!}&=\\[1.5ex]
\frac{(n-1)!(n-r+r)}{r!(n-r)!}&=\\[1.5ex]
\frac{n!}{r!(n-r)!}. 
\end{align*}
Which is $\binom{n}{r}$ by definition. 
A: You have made no mistake. You are just failing to see how this equals $\binom{n}{r}$.
Notice that every factor of $N_2$ is also a factor of $N_1$. And there is only one factor of $N_1$, that is not a factor of $N_2$, which is $(n-r)$.
You can write $N_1 = N_2 \times (n-r)$
Then, $\dfrac{N_1 + r\cdot N_2}{r!} = \dfrac{N_2 \times (n-r) + r\cdot N_2}{r!} = \dfrac{N_2 \times ((n-r)+r)}{r!} = \dfrac{N_2 \times n}{r!}$.
Now substitute for $N_2$ to see it equals $\binom{n}{r}$.
A: You have $n$ objects and you like to choose $r$ objects out of them.
Put one of them aside and choose your $r$ objects out of the remaining $n-1$ objects. 
That gives you $$\binom {n-1}{r}$$ choices.
Now put the object back into  the pile and choose that object and $r-1$ objects from the remaining $n-1$ objects, that gives you  $$\binom {n-1}{r-1}$$ choices.
That takes care of all your choices.
Thus $$\binom  {n}{r} =  \binom {n-1}{r} +\binom {n-1}{r-1}$$
