Tricky quantitative reasoning practice problem: Help

[jimmyjimjam]

Here's a tricky (in my opinion) practice problem from the Et Cetera part of the GRE test:

 

"An elected official wants to take five members of his staff to an undisclosed location.  What is the minimum number of staff members the elected official must have in order to have at least 20 different groups from which to choose?"

 

Now, I understand that is is a combination problem.  We need to know the least amount of staff members the official can have in order to be able to form 20 distinct groups of 5.  I understand that just because a member is in one group, doesn't preclude the possibility that they could serve in another group, making it a combination problem not a permutation problem.  The answer choices are:

 

a.  7

 

b.  8

 

c.  9

 

d.  10

 

e.  11

 

I know I need to plug in.  I'm just having a hard time visualizing what I need to plug in, and why I'm doing so.  Any help is much appreciated.

 

 

[saphixation]

So, the formula to find a combination is n choose k, or n! / (k! * (n - k)!) where n is the number of things to choose from (in this case, that's the number of staff members, so it's unknown) and k is the size of the group (in this case, k is 5). Knowing that, and knowing we need at least 20 groups, we can say:

n! / (5! * (n - 5)!) >= 20

It's possible to simplify this, but to be honest I would just start plugging in for n. We want to find the minimum n, so put in the smallest possible n first, which is 7. Now we have:

7! / (5! * (7-5)!) >= 20 which simplifies to

7! / (5! * 2!) >= 20

(7 * 6) / 2! >= 20

21 >= 20

This is true, so we know that 7 is the answer.

 

Hopefully this makes sense - I'm happy to try and explain more if needed! Also, sorry about the formatting; gradcafe is not the best place to write math equations.

[jimmyjimjam]

Cool.  Thanks a lot.  I'll try to use your method and see what I get.

[33andathirdRPM]

Yeppers. C(7,5) = 21.