What is $\dbinom{n}{n-1}$ for any positive integer $n$?  (Answer in terms of $n$.)
\begin{align*}
\dbinom{n}{n-1}&=\dfrac{n!}{(n-1)!~1!}\\
&=\dfrac{n\times(n-1)\times(n-2)\times(n-3)\times\cdots\times 2\times 1}{(n-1)\times (n-2)\times (n-3)\times \cdots \times 2\times 1}\\
&=\boxed{n}.
\end{align*}Also, $\binom{n}{n-1}$ is the number of ways to choose $n-1$ objects out of $n$.  This is equivalent to choosing $1$ object not to use.  Since there are $n$ different objects, there are $\boxed{n}$ ways to do this.
n
Level 2
Counting & Probability
Problem: What is $\dbinom{n}{n-1}$ for any positive integer $n$?  (Answer in terms of $n$.)
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find $\dbinom{n}{n-1}$, we start with the definition of the binomial coefficient. The binomial coefficient $\dbinom{n}{k}$ represents the number of ways to choose $k$ elements from a set of $n$ elements. The formula for the binomial coefficient is given by:

\dbinom{n}{k} = \frac{n!}{k!(n-k)!}

In this case, we have $k = n - 1$. Substituting $k$ into the formula gives:

\dbinom{n}{n-1} = \frac{n!}{(n-1)!(n - (n-1))!} = \frac{n!}{(n-1)! \cdot 1!}

Next, we simplify the expression. The factorial $n!$ can be expressed as $n \cdot (n-1)!$. Thus, we have:

\dbinom{n}{n-1} = \frac{n \cdot (n-1)!}{(n-1)! \cdot 1} = \frac{n \cdot (n-1)!}{(n-1)!} = n

This simplification shows that the number of ways to choose $n-1$ elements from a set of $n$ elements is equal to $n$. Therefore, the final answer is $n$.<|END_OF_LATENT|> The final answer is $n$.