Intuitive explanation of binomial coefficient formula Regarding the formula for binomial coefficients:
$$\binom{n}{k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$$
the professor described the formula as first choosing the $k$ objects from a group of $n$, where order matters, and then dividing by $k!$ to adjust for overcounting.
I understand the reasoning behind the numerator but don't understand why dividing by $k!$ is what's needed to adjust for overcounting. Can someone please help me understand how one arrives at $k!$? Thanks.
 A: Suppose you had 5 books on a shelf and wanted to pick 3. If order mattered (permutations), there would be $5 \cdot 4 \cdot 3$ ways of doing so. If order did not matter (combinations), you would discount the extra permutations. That is, "ABC" would be the single representative of all of its $3!$ permutations. The total number ways of selecting the books is $\frac{5\cdot4\cdot3}{3!}$. For the general case, replace the 5 with $n$ and 3 with $k$.
A: Suppose you want to plant three identical red flowers in a row. In how many ways can you do it? First you plant all three red flowers in a row. Suppose you swap the first flower with the second one. Did anything change? You still have three red flowers in a row. Try swapping the first flower with the third one. Do you see any difference? I can't. They are still the same three red flowers planted in a row. Basically, no matter how your permute these flowers, you have only one outcome, namely a row of three identical red flowers. 
So, there are $3!$ ways to plant these flowers in a row, but since every row looks exactly like the other row, we divide out the redundant ones. There are $3!$ redundant rows. Then $\frac {3!}{3!} = 1$ way to plant three identical red flowers. 
A: Let's try it with $\dbinom 5 3:$
\begin{align}
P(5,2) = {} & (\text{number of permutations of 3 things in a set of 5}) \\[10pt]
= {} & \frac{5!}{(5-2)!} = 20. \\[10pt]
& \text{These permutations are:} \\[10pt]
& ab,\ ba\qquad \longleftarrow \text{$2!=2$ permutations, both the same combination.} \\
& ac,\ ca\qquad \longleftarrow \text{$2!=2$ permutations, both the same combination.} \\
& ad,\ da\qquad \longleftarrow \text{$2!=2$ permutations, both the same combination.} \\
& ae,\ ea\qquad \longleftarrow \text{etc.} \\
& bc,\ cb \\
& bd,\ db \\
& be,\ eb \\
& cd,\ dc \\
& ce,\ ec \\
& de,\ ed
\end{align}
For each combination, there are $2!=2$ permutations. So you divide the number of permutations by $2!=2$ to get the number of combinations.
