Need HELP with this Difficult Kaplan GMAT math question

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

relaxin99: Senior | Next Rank: 100 Posts; Posts: 73; Joined: Tue Sep 16, 2008 4:33 pm; Thanked: 2 times

Need HELP with this Difficult Kaplan GMAT math question

by relaxin99 » Tue Sep 16, 2008 5:00 pm

Hi just started taking the Kaplan gmat course and we took a full length exam, well i had a problem with this problem solving question. Can someone PLEASE HELP

A local musem is setting up a new exhibit. 5 oil paintings and 6 watecolor paintings are available for the exhibit. The museum director must choose 3 oil paintings and 4 watercolors. How man different bombinations of paintings for the exhibit are possible?

A) 12
B) 30
C) 90
D) 120
E) 150

the answer is E)150 but i have NO idea how to go about working it, can someone please work this problem out for me, greatly appreciated.

Quote

Source: Beat The GMAT — Problem Solving | Tue Sep 16, 2008 5:00 pm

GG04: Senior | Next Rank: 100 Posts; Posts: 51; Joined: Fri Sep 05, 2008 11:06 am; Thanked: 1 times

by GG04 » Tue Sep 16, 2008 5:32 pm

This is a regular Permutation Combination problem
Out of 5 oil paintings, he chooses 3. that is 5C3 = 5!/2!3! = 10
Out of 6 water color, he choose 4 = 6!/4!2! = 15

Since he has to choose both oil and water color, the answer is 15 * 10 = 150

Quote

cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pm; Thanked: 331 times; Followed by:11 members

by cramya » Tue Sep 16, 2008 10:07 pm

Relaxin,

Choosing k things from n things available (usually involves either permutations or combinations)

Here the order does not matter. Why? Lets say there are 5 paintings
A,B,C,D,E and I have to choose 3. If I choose A,B,D its the same as me choosing B,A,D or D,A,B or D,B,A or A,D,B or B,D,A.This applies to any set of 3 choosen

So its C(5,3) = n C r = 5!/3!*2! = 10

Silimarly for the 6 water colors its 6C4 = 15

Now I have 10 different combinations of paint and 15 different combinations of water colors i.e. choosing the paint can happen in 10 different ways(lets say m) and each of these m ways can be followed by choosing the watercolor in 15 different ways (lets say n) then by the fundamental counting principle the 2 events can happen in m*n ways (i.e 10*15 = 150)

Hope this helps!

Quote

tendays2go: Senior | Next Rank: 100 Posts; Posts: 55; Joined: Sun Sep 14, 2008 6:51 am; Location: Netherlands; Thanked: 10 times; GMAT Score:680

by tendays2go » Tue Sep 16, 2008 11:58 pm

Another alternate expln:

This is a combination problem not a permutation one, as the order of paintings here is not important.
i.e. in the exhibition, there is no mention of ordering of the paintings, or there's no restriction/ condition either.
So, it's a combination problem.

From now, on it's straighfwd:
the valid combinations are: 5C3 * 6C4 = (5!/(3!2!)) * (6!/(4!2!))
= 10 * 15 = 150

Quote

sudi760mba: Senior | Next Rank: 100 Posts; Posts: 85; Joined: Wed Aug 13, 2008 2:21 am; Thanked: 1 times

by sudi760mba » Wed Sep 17, 2008 3:51 am

Permutation: order matters n!/(n-k)!
Combination: order does not matter n!(n-k)!k!

Combination problem 5!/(5-3)!3! = 5!/2!*3! = 3*4*5/3 *2 = 60/6 = 10
6!/(6-4)!4! = 6!/2!*4! = 3*4*5*6/4*3*2 = 360/24 = 15

In this case, the 2 cancels below leaving 3*4*5*6 in the numerator to be divided by 4 * 3 * 2 in the denominator

or alternatively
6!/2!4! = 5 * 6 /2 *1 = 30/2=15

In this case, the 4 cancels and that leaves 5 * 6 in the numerator to be divided by 2 factorial or 2 = 15

10 * 15 = 150 possible combinations