number of vacancies

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shibal: Master | Next Rank: 500 Posts; Posts: 345; Joined: Wed Mar 18, 2009 6:53 pm; Location: Sao Paulo-Brazil; Thanked: 12 times; GMAT Score:660

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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Source: Beat The GMAT — Problem Solving | Thu Apr 23, 2009 6:50 pm

dmateer25: Community Manager; Posts: 1049; Joined: Sun Apr 06, 2008 5:15 pmFollowed by: 27 members; Location: Pittsburgh, PA; Thanked: 113 times; GMAT Score:710

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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shibal: Master | Next Rank: 500 Posts; Posts: 345; Joined: Wed Mar 18, 2009 6:53 pm; Location: Sao Paulo-Brazil; Thanked: 12 times; GMAT Score:660

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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dmateer25: Community Manager; Posts: 1049; Joined: Sun Apr 06, 2008 5:15 pmFollowed by: 27 members; Location: Pittsburgh, PA; Thanked: 113 times; GMAT Score:710

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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shibal: Master | Next Rank: 500 Posts; Posts: 345; Joined: Wed Mar 18, 2009 6:53 pm; Location: Sao Paulo-Brazil; Thanked: 12 times; GMAT Score:660

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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rossmj: Senior | Next Rank: 100 Posts; Posts: 60; Joined: Thu Feb 26, 2009 12:50 pm; Location: Chicago; Thanked: 3 times

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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dendude: Master | Next Rank: 500 Posts; Posts: 160; Joined: Fri May 30, 2008 7:10 pm; Thanked: 10 times; GMAT Score:600

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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