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number of vacancies

by shibal » Thu Apr 23, 2009 6:50 pm
to fill a number of vacancies, an employer must choose 3 programmes from 6 participants and 2 managers from 4 applicants. what is the total number of ways in whihc she can make her selection?

OA 120 , i get 132 , what i'm doing wrong?
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by dmateer25 » Thu Apr 23, 2009 7:22 pm
6C3 x 4C2

6C3 = 6!/3!3! = 20

4C2 = 4!/2!2! = 6

20 * 6 = 120
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by shibal » Fri Apr 24, 2009 4:42 am
to get the programmers doesn't she have 120 different options (6*5*4)??

when you use 6C3 you mean the formula n!/(n-k)! ?
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by dmateer25 » Fri Apr 24, 2009 5:53 am
shibal wrote:to get the programmers doesn't she have 120 different options (6*5*4)??

when you use 6C3 you mean the formula n!/(n-k)! ?
I mean the formula n!/(n-k)! k!


You can't do 6 x 5 x 4 because you are counting all of the duplicates.

For example, lets label the 6 people: A B C D E F

Now lets say the group is picked as: A B C. Then the next group is picked as C B A. This is actually the same group and you are double counting it if you do 6 x 5 x 4.

Thus, you have to use the combination formula to eliminate these duplicates.

Again, this formula is n!/(n-k)! k!.

So for choosing 3 people from 6. You would calculate:

6!/3! (6-3)! = 6!/3!3! = (6 x 5 x 4)/ (3 x 2) = 5 x 4 = 20
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by shibal » Sat Apr 25, 2009 4:59 am
dmateer25 (or anyone else) what is the best way to figure out when to use n!(n-k)!k! and n!(n-k)! ?
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by rossmj » Mon Apr 27, 2009 7:11 am
The best way to determine is to see if order matters. If we need to select a commitee of 3 people and 6 people have been nominated then we would use the 6!/3!(6-3)! because a group of Tim, Tom, and Tammy is the same thing as a group of Tammy, Tim and Tom.

If we had to select 3 people from 6 to fill the roles of President, Vice President, and Secratary we would use 6!/(6-3)! because President Tim, VP Tom, and Sec. Tammy is different from Pres. Tammy, VP Tim, and Sec. Tom.
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by dendude » Mon Apr 27, 2009 9:23 am
shibal wrote:dmateer25 (or anyone else) what is the best way to figure out when to use n!(n-k)!k! and n!(n-k)! ?
You are in turn asking for the difference between a combination and a permutation.
nCk = n!/(n-k)!*k!
nPk = n!/(n-k)!

Combination refers to a Selection of things from a group.
Eg. Choosing 4 books out of a pile of 10 books.

Permutation refers to arranging things from a group
Eg. Arranging 4 books on a shelf out of a pile of 10 books.

Order matters in Permutations (i.e 1,2,3 diff from 3,2,1), hence P > C (for same n and k, ofcourse)
Think of it in this way, essentially P and C are the same but you need to lessen C by a factor of k! to eliminate the duplicates.
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