Binomial Theorem Identities What's the actual difference between these two formulas (they're both in the chapter regarding binomial theorem).  They're from two different textbooks :
$${n\choose k}+{n\choose k+1}={n+1\choose k+1}$$
and 
$${n-1\choose k}+{n-1\choose k-1}={n\choose k}$$
I'll be appreciated if someone explain it to me either combinatorially or algebraically.  Thanks!
 A: There’s no real difference: they express the same fact about binomial coefficients. To see this, first replace $n$ by $m$ and $k$ by $\ell$ in the second equation to get
$$\binom{m-1}\ell+\binom{m-1}{\ell-1}=\binom{m}\ell\;.\tag{1}$$
Now let $\ell=k+1$ and $m=n+1$; then $(1)$ becomes
$$\binom{n}{k+1}+\binom{n}k=\binom{n+1}{k+1}\;,$$
which is your first equation.
In terms of Pascal’s triangle each of these equations says that each entry is the sum of the two above it.
A: $$\binom n k+\binom n{k+1}$$
$$=\frac {n!}{k!(n-k)!}+\frac{n!}{(k+1)!(n-k-1)!}$$
$$=\frac{n!}{(k+1)! (n-k)!}\left(k+1+n-k\right)$$
$$=\frac{(n+1)!}{\{(n+1)-(k+1\}!)(k+1)!}$$
$$=\binom{n+1}{k+1}$$
A: We simply use the definition of Combination $${n \choose k} = {n! \over {(n-k)!k!}}$$
The top one goes like this $${n \choose k} + {n \choose k+1}$$
$$= {n! \over {(n-k)!k!}} +{n! \over {(n-(k+1))!(k+1)!}}$$
$$= {{n!(k+1) }\over {(n-k)!k!(k+1)}} +{{n!(n-k)} \over {(n-k)!k!(k+1)}}$$
$$={n![(k+1)+(n-k)]\over{(n-k)!(k+1)!}}$$
$$={(n+1)!\over{(n-k)!(k+1)!}}$$
$$={ {n+1} \choose {k+1}}$$
The bottom one can be proven using the top one just by substitution :)
A: Combinatorial proof of Pascal identity
Denote by $$C^k(I_n)=\{A:A\subset I_n,|A|=k\},|C^k(I_n)|=\binom{n}{k}$$ the set of all k-subsets of set $I_n=\{0,1,2,...,n-2,n-1\}$ and then define a set 
$$C'=\{A:A=C_1\cup\{n-1\},C_1\in C^{k-1}(I_{n-1})\}$$
now is clear that
$$|C'|=|C^{k-1}(I_{n-1})|=\binom{n-1}{k-1}$$
because
$$C^k(I_n)=C'\cup C^k(I_{n-1})$$
$$C'\cap C^k(I_{n-1})=\emptyset$$
follow that
$$|C^k(I_n)|=|C'|+|C^k(I_{n-1})|$$
or
$$\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$$
