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I am working with a combination and have the problem $_{v}C_{v-2} = \binom{v}{v-2} $ I know the combination formula but can't figure out how to simplify my answer to get the correct answer. How do i do this? I can't seem to get the right answer.

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$$\begin{aligned} \binom{v}{v-2} & = \frac{v!}{(v-2)!(v - (v-2))!} \\ &= \frac{v!}{2(v-2)!} \\ &= \frac12 \frac{v \cdot (v - 1) \cdot \color{blue}{(v - 2)} \cdot \color{green}{(v - 3)} \; \color{red}{\cdots} \; \color{purple}{2} \cdot \color{darkorange}{1}}{\color{blue}{(v-2)} \cdot \color{green}{(v-3)} \; \color{red}{\cdots} \; \color{purple}{2} \cdot \color{darkorange}{1}} \end{aligned}$$

What do you do now? (Hint: I've colored it for you.)

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I see that i can cancel out the v-2 and the v-3. But how did you get there in the first place? – Jessica Apr 17 '13 at 21:58
@Jessica, not only that, you can cancel out all the terms (note the $\cdots$ at the end). Since $n! = n(n-1)(n-2) \cdots 2 \cdot 1$ (by definition), the result follows. – George V. Williams Apr 17 '13 at 22:00
thank you I understand it now! – Jessica Apr 17 '13 at 22:03
@Jessica, you can tick the checkmark to the left of this question to indicate to others that this question has been answered. – George V. Williams Apr 17 '13 at 22:09

Note that by symmetry we have:


Use this to eliminate one of the variables.

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The number of ways to select $v-2$ out of $v$ things and put them into a box is the same as the number of ways to select $2$ out of $v$ things not to put into the box.

Or: The coefficient of $x^{v-2}$ in $(1+x)^v$ is the same as the coefficient of $x^2$ in $(x+1)^v$.

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