How many ways can you choose a commitee of 3 people from a group of 12?

Question

a. 3
b. 36
c. 12
d. 72
e. 220

Show your work and explain.

Answer 1

C(12,3)
That is, 12!/((3!x9!) =
(12x11x10)/(3x2) = 220 (e)
You use combination rather than permutation because on a committee order doesn't matter.
Remember combination = committee permutation = position
for more info search on permutations combinations math
You'll turn up stuff like this...
http://www.mathgoodies.com/forums/topic.asp?ARCHIVE=true&TOPIC_ID=30541
http://mathforum.org/dr.math/faq/faq.comb.perm.html
Dusty Alfo's answer is a permutation, not a combination. That's how many ways you could choose 3 people out of 12 and assign to certain positions, for example, president vice president secretary. The reason why you'd have to have the extra 3! in the bottom is because choosing person C, then perosn B, then person A, for a committee, is the same as choosing preson A, then person B, then person C. 3! is the number of ways you can arrange these three people once you've chosen them, and you need to divide this out of the combination because the ways these three people are arranged doens't matter.

Answer 2

There are two different types of problems that are easily confused with this type of information, called permutations and combinations. The difference is, does the order matter?

In this example, the order that we choose the three people for the commitee does not matter. This means that if Adam, Bob, and Chris get chosen, it doesn't matter who gets picked first, second or third. Order would matter if they said that the commitee would have a president, vice president, and treasurer.

To start with, we'll find out how many ways 12 people can fill 3 positions. We have 12 choices for the first person in the commitee, then 11 people to choose from for the second slot, and 10 people left to take the third spot.

12 x 11 x 10 = 1320

However, this does not eliminate the numerous different ways that the same three people could be picked. If we go back to Adam, Bob, and Chris from above, notice that the three of them could make the commitee in the following six orders: ABC, ACB, BAC, BCA, CAB, or CBA. All six ways would give you the exact same committe, so we need to eliminate those duplicates.

1320/6 = 220.

In short, for all problems like this where order doesn't matter, find the combinations possible, then divide by the number of duplicates.

(12 x 11 x 10) / (3 x 2 x 1) = 220



If the problem changed to a committee of 4 unique positions chosen from a group of 10, it would look like this:

(10 x 9 x 8 x 7) = 5040

But, if the committee didn't have unique positions and simply had 4 members, you'd have to get rid of the duplicate committees by dividing.

(10 x 9 x 8 x 7) / (4 x 3 x 2 x 1) = 210.

As you can see, it makes a big difference. Easy rule to remember this:

If order does matter, do not divide.
If order does not matter, do divide.

Answer 3

To choose 3 people from a group of 12, perform the operation 12C3, or C(12,3), or "12 choose 3." The formula is nCr = n! / r!(n - r)!, where ! is the factorial operation. 12C3 = 12! / 3!9! = (1*2*3*4*5*6*7*8*9*10*11*12) / (1*2*3)(1*2*3*4*5*6*7*8*9) = 10*11*12 / 1*2*3 = 1320 / 6 = 220.

Answer 4

The solution is 12!/((12-3)!*3!) or 12*11*10/(3*2*1) = 220

The division by 3! is because it doesn't matter what order the committee members are chosen.

Answer 5

C(12,3) =12*11*10/ (3*2*1) =220
Answer = (e)

Answer 6

This is calculated by the combination of 3 from 12, which is

12!/(3!(12-3)!) = 220

Where ! means factorial

Answer 7

put this in your calculator, 12 then go to math, probability, then scroll down to nCr, then hit enter, and then put in 3 and you have your answer

Answer 8

12!/(3!*(12-3)!)=12!/(3!*9!)=220

e. 220 is correct

Answer 9

it would be 12x11x10 because you select a person, then you have 11 left to choose, then you select the 2nd person, and have 10 to choose from. it should be 1320 ways, if im not mistaking. please email me or something and tell me if its wrong.

Answer 10

12C3 = (12!)/(3!(12 - 3)!)
12C3 = (12!)/(3! * 9!)
12C3 = 220

ANS : e.) 220