How to do mathematical permutations/combinations?

Question

Here is the question, I do not know the formula for figuring out combinations:

An executive director wants to take a group of 5 of his employees to a business retreat. What is the minimum amount of staff members he needs to have to make at least 20 different groups to choose from?

a)7 b)8 c)9 d)10 e)11

The answer to this question is 7, but I do not know how I am to arrive at that conclusion, will someone familiar with the equation please help. :)

Thank you all, very great answers. I think I have it, but let me put the combinations formula in layman's terms to be sure...

total sample size factored out divided by amt chosen factored X the amt left factored...

7!/5!2! ???

Answer 1

Find 7C5, which is called a combination, "7 choose 5"

The formula is 7!/[5!2!] = 5040/240 = 21.

The executive director can make 21 groups when choosing 5 from 7.

In general, the formula for a combination, when choosing r
objects from a set of n, is:

n!/[(n-r!)(r!)]

In this case it looks like it would have been best to start at 7, and use the formula to see when you would get 20 or more.

Answer 2

try pascals triangle
Row 1 1
Row 2 1 1
Row 3 1 2 1
Row 4 1 3 3 1
Row 5 1 4 6 4 1
Row 6 1 5 10 10 5 1
Row 7 1 6 15 20 15 6 1

also this site

http://mathforum.org/dr.math/faq/faq.comb.perm.html

good luck

Answer 3

Take a calculated guess at 7 and get 7C5=21 which >20.

To solve for n directly means solving a fifth degree equation.

Answer 4

All of these answers are correct, but just FYI it is good to know the difference between a permutation and a combination -- otherwise you won't know how to approach these problems.

In permutations ORDER MATTERS. Things like lottery tickets or batting lineups in baseball are permutations

In combinations ORDER DOESN'T MATTER. a 5 card hand in poker, or a comittee or group are combinations.

Hope that helps

Answer 5

nCr = n!/[(n-r)!*r!]

or xC5 = 20

it would be best to guess if given multiple choice