Evaluate a sum involving n choose r Evaluate: $\sum_{k=0}^6 (-1)^k \binom{6}{k}$ where $\binom{n}{r}= \frac{n!}{r!(n-r)!}$.
I'm unsure how to compute the part with $\binom{6}{k}$, it should be something along the lines of $\frac{6!}{k!(6-k)!}$ can we simplify this down? 
I think there was an expansion method for factorials I could used in the denominator that was somewhere along the lines of $(6-k)(5-k)(4-k)(3-k)(2-k)(1-k)$ ?
 A: The most transparent way to look at this is to use the binomial theorem, as both earlier answers suggest,
$$
(p+q)^n=\sum_{k=0}^n\binom{n}{k}p^{n-k}q^k
$$
and then set $n=6$, $p=1$, $q=-1$.
A: Hints:
$$nCr:=\binom{n}r\,,\;\;\text{and by the Binomial Theorem (Newton), we get :}$$
$$(a+b)^n=\sum_{k=0}^n\binom nka^{n-k}b^k$$
Substitute above now with $\,a= 1 \;,\;b = -1\;,\;n=6$ ...
A: Hint: The binomial theorem tells that that: $$(x+y)^n=\sum_{i=0}^n\left(\begin{array}{c} n \\ i \end{array}\right)x^iy^{n-i}$$where $\left(\begin{array}{c} n \\ i \end{array}\right)$ is your nCi. Maybe some clever choices for $x$, $y$, and $n$ will make your summation easier.
A: Since the question has been answered a few times, let's answer a more general question of the same form.
$$
\begin{align}
\sum_{j=k}^m(-1)^j\binom{n}{j}
&=\sum_{j=k}^m(-1)^j\left(\binom{n-1}{j}+\binom{n-1}{j-1}\right)\\
&=\sum_{j=k}^m(-1)^j\binom{n-1}{j}+\sum_{j=k-1}^{m-1}(-1)^{j+1}\binom{n-1}{j}\\
&=(-1)^m\binom{n-1}{m}+(-1)^k\binom{n-1}{k-1}
\end{align}
$$
The third equality takes advantage of the term by term cancellation for $j=k$ throught $j=m-1$. This leaves only the $j=m$ term on the left and the $j=k-1$ term on the right.
In the particular case at hand, $k=0$ and $m=n=6$. Since $\displaystyle\binom{5}{6}=\binom{5}{-1}=0$, the sum is $0$.
