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In how many ways can 3 different portraits be mounted

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by Vincen » Sat Apr 07, 2018 1:33 am
M7MBA wrote:In how many ways can 3 different portraits be mounted on 3 of the four walls inside a room???

A) 4
B) 6
C) 12
D) 16
E) 24

The OA is the option E .

It shouldn't be 9? 3 walls and 3 portraits, 3*3=9. No? I need some help here. Thanks in advanced.
Hello M7MBA.

Let's see, we have:

- 4 walls from which we have to pick 3, it can be done as follows <i class="em em-point_down"></i> $$4C3\ =\frac{4!}{3!1!}=4\ different\ ways.$$

- We need to arrange 3 portraits, this can be done in 3! different ways, that is to say, 6 different ways.

Therefore, the number of ways that 3 different portraits can be mounted on 3 of the four walls inside a room is given by <i class="em em-point_down"></i> $$4\cdot3!=4\cdot6=24\ different\ ways.$$ This implies that the correct answer is the option E .

I hope it can help you. <i class="em em---1"></i>
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by GMATGuruNY » Sat Apr 07, 2018 2:15 am
M7MBA wrote:In how many ways can 3 different portraits be mounted on 3 of the four walls inside a room???

A) 4
B) 6
C) 12
D) 16
E) 24
Number of options for the first portrait = 4. (Any of the 4 walls.)
Number of options for the second portrait = 3. (Any of the 3 remaining walls.)
Number of options for the last portrait = 2. (Either of the 2 remaining walls.)
To combine the options above, we multiply:
4*3*2 = 24.

The correct answer is E.
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by Scott@TargetTestPrep » Thu Jul 05, 2018 3:53 pm
M7MBA wrote:In how many ways can 3 different portraits be mounted on 3 of the four walls inside a room???

A) 4
B) 6
C) 12
D) 16
E) 24
Solution:

We have 4C3 = 4 ways to choose the 3 of the 4 walls. Next we have 3! = 6 ways to put the portraits onto the 3 walls we've chosen. Thus, there are 4 x 6 = 24 ways to mount these portraits.

Answer: E

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