# How many ways can the school choose a President Vice President?

There are n >= 4 students. The school has a Board of Directors, consisting of one president and three vice-presidents. The entire board consists of four distinct students. How can I prove that $$n\binom{n-1}{3} = (n-3)(\binom{n}{3})$$ by couting in two different ways, the number of ways to choose a board of director?

I know that out of 4 student there is 4 way to choose a president because anyone of the students can be choosen, which leave 3 students and out of those three student there is only 1 way of choosing a vice-president because the board consit of three vice-president. I can see that the equation also equals to 4=4 but how do i prove it by counting in two different ways?

First choose the vice presidents. There are $n$ selections for the first choice, $n-1$ for the second, and $n-2$ for the third. Since these are all taken in sequence, we multiply these numbers. Since the order of the group of three that we choose doesn't matter (there isn't a first-place vice president, second-place vice president, third-place vice president, in other words), we then cut out the number of ways we can arrange the group of three of our choice, leaving us $$\frac{n(n-1)(n-2)}{3*2*1}=\frac{n!}{(n-3)!3!}=\binom{n}{3}$$ ways to choose three vice presidents.

There are now $n-3$ people remaining, from which we must now choose a president. Putting this all together, we have $$(n-3)\binom{n}{3}=\frac{n(n-1)(n-2)(n-3)}{3*2*1}$$ ways to choose one president and three vice presidents.

Reverse the order in which you choose, i.e. now choose one president first, meaning $n$ choices, then from the group of $n-1$ remaining, make three choices (again, the order doesn't matter for the choices since the vice presidents are not distinguished from one another). You then have $$(n)\binom{n-1}{3}=n*\frac{(n-1)!}{(n-3-1)!3!}$$ ways to choose a president followed by three vice presidents. I will leave the final details to you to show that the above equation is equal to our first total.

Both sides of the equation are equal to the number of ways you can select a President and Vice Presidents as described.

For the left side, if you choose one of the students to be president first, there are $n$ possibilities. After this, you must choose 3 students from the remaining $n-1$ to be vice presidents, which gives $\binom{n-1}{3}$ possibilities. All up this gives

$$n\binom{n-1}{3}$$

ways of choosing a President and Vice Presidents.

For the right side, if you choose the three vice presidents first, from the $n$ students, there are $\binom{n}{3}$ ways of doing this. Then all that remains is to choose the president from the remaining $n-3$ students, for which there are $n-3$ possibilities. In total, this gives

$$(n-3)\binom{n}{3}$$

ways once again, of choosing a President and Vice Presidents.

Therefore $$n\binom{n-1}{3}=(n-3)\binom{n}{3}.$$