0
$\begingroup$

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.$ $11$

$\endgroup$
8
  • $\begingroup$ How can I set up a formula? $\endgroup$
    – user268238
    Sep 5 '15 at 17:39
  • 3
    $\begingroup$ 5 minutes of his staff? $\endgroup$
    – spyr03
    Sep 5 '15 at 17:39
  • 1
    $\begingroup$ Question needs to be a little more clear.. $\endgroup$
    – Soham
    Sep 5 '15 at 17:40
  • 1
    $\begingroup$ Are they going fishing for red herrings in that undisclosed secure location? $\endgroup$ Sep 5 '15 at 17:45
  • 1
    $\begingroup$ @user268238 If there are $x$ people, how many ways are there to choose a group of 5 members? $\endgroup$ Sep 5 '15 at 17:54
0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ This is none of my answer choices. 2400 is way off. $\endgroup$
    – user268238
    Sep 5 '15 at 18:00
  • $\begingroup$ Try plugging $x=\text{one of your choices}$ into the polynomial shown and see if the answer is bigger than $2400$. $\endgroup$
    – Will R
    Sep 5 '15 at 18:04
  • $\begingroup$ All of the x vales in my choices work, and the smallest one is x=7. So the answer choice is A $\endgroup$
    – user268238
    Sep 5 '15 at 18:16
  • $\begingroup$ @user268238 pefect, and for completeness, show that 6 is too small, and you are done $\endgroup$
    – spyr03
    Sep 5 '15 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.