Jack is making a list of his 5 favorite cities. He will choose 3 cities

[BTGModeratorVI]

Jack is making a list of his 5 favorite cities. He will choose 3 cities in the United States from a list of 5 candidates. He will choose 2 cities in Europe from a list of 3 candidates. How many different lists of cities, ranked from first to fifth, can Jack make?

A. 30
B. 360
C. 1,800
D. 3,600
E. 6,720

Answer: D
Source: Princeton Review

[Brent@GMATPrepNow]

Jack is making a list of his 5 favorite cities. He will choose 3 cities in the United States from a list of 5 candidates. He will choose 2 cities in Europe from a list of 3 candidates. How many different lists of cities, ranked from first to fifth, can Jack make?

A. 30
B. 360
C. 1,800
D. 3,600
E. 6,720

Answer: D
Source: Princeton Review

Take the task of creating a list and break it into stages.

Stage 1: Select 3 US cities
Since the order in which we SELECT the cities does not matter, we can use combinations.
We can select 3 cities from 5 cities in 5C3 ways (10 ways)
So, we can complete stage 1 in 10 ways

Stage 2: Select 2 European cities
Since the order in which we SELECT the cities does not matter, we can use combinations.
We can select 2 cities from 3 cities in 3C2 ways (3 ways)
So, we can complete stage 2 in 3 ways

Stage 3: Arrange the 5 selected cities
We can arrange n unique objects in n! ways.
So, we can arrange the 5 selected cities in 5! ways (120 ways)
We can complete this stage in 120 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create our list) in (10)(3)(120) ways (= 3600 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch this free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

[Scott@TargetTestPrep]

Jack is making a list of his 5 favorite cities. He will choose 3 cities in the United States from a list of 5 candidates. He will choose 2 cities in Europe from a list of 3 candidates. How many different lists of cities, ranked from first to fifth, can Jack make?

A. 30
B. 360
C. 1,800
D. 3,600
E. 6,720

Answer: D
Source: Princeton Review

Jack can choose the 3 cities in the U.S. in 5C3 = 5!/(3!*2!) = (5 x 4)/2 = 10 ways.

He can choose the 2 cities in Europe in 3C2 = 3 ways.

Thus, in total, the five cities can be chosen in 3 x 10 = 30 ways. Since he will also order the cities, he can make 30 x 5! = 30 x 120 = 3,600 lists.

Answer: D

[deloitte247]

No of ways to choose three American cities from a list of five
$$5C_3$$

No of ways to choose two European cities from a list of three
$$3C_2$$

No of unique ways to arrange the 5 cities = 5!

Total no of possible list =
$$\left(5C_3\right)\left(3C_2\right)\left(5!\right)$$
$$\left(\frac{5!}{3!\left(5-3\right)!}\right)\left(\frac{3!}{2!\left(3-2\right)!}\right)\left(5!\right)$$
$$\left(\frac{5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1\left(2\cdot1\right)}\right)\left(\frac{3\cdot2\cdot1}{2\cdot1\left(1\right)}\right)\left(5\cdot4\cdot3\cdot2\cdot1\right)$$ $$\left(10\right)\left(3\right)\left(120\right)=3600\ possible\ lists$$
$$Answer\ is\ Option\ D$$