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

Question

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

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

BrentGMATPrepNow · Accepted Answer

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

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pushpitkc · Answer

Ways of choosing his top 3 american cities from a list of 5: 5c3
Ways of choosing his top 2 american cities from a list of 3: 3c2
Ways of arranging them in order from 1 to 5: 5!

5c3 * 3c2 * 5! 
= 10 * 3 * 120
= 3600(Option D)

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Kritesh · Answer

The selection process in the United states is 5C3 and that in Europe is 3C2. (Since its a selection process we use combinations).

Therefore we get   = 30 ---- Upon simplification. (1)

The Trap here would be to consider the above as the answer.

At the end of the statement it says that  "different lists of cities, ranked from first to fifth" .

Therefore the Cities can be arranged in 5! Ways which is 5*4*3*2*1 = 120 (2)

From (1) and (2) we get 30*120 = 3600.

Therefore option D is the answer.

ScottTargetTestPrep · Answer

If the ranking doesn’t matter, there are 5C3 x 3C2 lists of 5 cities. However, since, for each list of the 5 cities, there are 5! ways to rank the cities from first to fifth, the total number of ranked lists is:

5C3 x 3C2 x 5!

(5 x 4 x 3)/(3 x 2) x 3 x 120

10 x 3 x 120 = 3,600

Answer: D

Fdambro294 · Answer

(1st)Needs to Choose which 3 American Cities and which 2 European Cities will be ranked.

Out of the 5 possible American cities, he can choose Groups of 3 in:

"5 choose 3" = Number of Ways

AND

For Each Combination of 3 Unique American Cities chosen, he then picks 2 out of 3 European Cities. He does this in:

"3 choose 2" = Number of Ways

All the Different Combinations of 3 Unique American Cities and 2 Unique European Cities that can be chosen is found by Multiplying:

(5! / 2!3!) * (3! / 2!1!) = 30 Different Ways to Choose

AND

(2ND) For Each Unique Combination of 5 Cities chosen, he must Arrange the Cities from Least - to - Worst. If there is N Unique Elements, they can be Arranged in N! Ways.

5! = 120

Answer:

(30) * (120) = 3,600

-D-

Archit3110 · Answer

there are 3 parts to the question: 
1. ways to choose 3 cities from 5 United states : 5c3 ;10
2. 2 cities from list of 3 Europe ; 3c2; 3
3. Arrangement list of 5 cities in different ways ; 5! ; 120
total ways ; 10*3*120 ; 3600
option D

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

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

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