# Permutations and combinations

An elected staff member wants to take five members of his staff to an undisclosed secure location. How many members must the elected official employ in order to have a minimum of 20 different groups from which to choose?

Choices:

$A.$ $7$

$B.$ $8$

$C.$ $9$

$D.$ $10$

$E.$ $11$

• How can I set up a formula? Sep 5 '15 at 17:39
• 5 minutes of his staff? Sep 5 '15 at 17:39
• Question needs to be a little more clear.. Sep 5 '15 at 17:40
• Are they going fishing for red herrings in that undisclosed secure location? Sep 5 '15 at 17:45
• @user268238 If there are $x$ people, how many ways are there to choose a group of 5 members? Sep 5 '15 at 17:54

Suppose he has $x$ people to choose from. Then we know that he can make $x(x - 1)(x - 2)(x - 3)(x - 4) / 5!$ groups, since permutation does not matter.
In other words, he can pick one person first, and he will have $x-1$ people left to choose from for the next person. He does this $5$ times, hence the formula.
You may have seen it written as $\binom{n}{k} = \frac{n!}{k!(n-k)!}$
So the question is when is \begin{eqnarray*} {1 \over 5!} \cdot x(x - 1)(x - 2)(x - 3)(x - 4) & \geq & 20\\ x(x - 1)(x - 2)(x - 3)(x - 4) & \geq & 5! \cdot 20\\ x(x - 1)(x - 2)(x - 3)(x - 4) & \geq & 2400 \end{eqnarray*} And since $x$ is an natural number, i.e., $\{0, 1, 2, 3, 4, ...\}$ you can just try answers until you hit it.
• Try plugging $x=\text{one of your choices}$ into the polynomial shown and see if the answer is bigger than $2400$. Sep 5 '15 at 18:04