Veritas Prep
Anna has to visit at least 2 European cities on her vacation trip. If she can visit only London, Paris, Rome, or Madrid, how many different itineraries, defined as the sequence of visited cities, can Anna create?
A. 12
B. 36
C. 48
D. 60
E. 72
OA D
Anna has to visit at least 2 European cities on her vacation
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We need to consider 3 different cases:AAPL wrote:Veritas Prep
Anna has to visit at least 2 European cities on her vacation trip. If she can visit only London, Paris, Rome, or Madrid, how many different itineraries, defined as the sequence of visited cities, can Anna create?
A. 12
B. 36
C. 48
D. 60
E. 72
OA D
i) Anna visits 2 cities
ii) Anna visits 3 cities
iii) Anna visits 4 cities
i) Anna visits 2 cities
There are 4 options for the FIRST city and 3 options for the SECOND city
So, the total number of 2-city itineraries = (4)(3) = 12
ii) Anna visits 3 cities
There are 4 options for the FIRST city, 3 options for the SECOND city, and 2 options for the THIRD city
So, the total number of 3-city itineraries = (4)(3)(2) = 24
ii) Anna visits 4 cities
There are 4 options for the FIRST city, 3 options for the SECOND city, 2 options for the THIRD city, and 1 option for the FOURTH city
So, the total number of 4-city itineraries = (4)(3)(2)(1) = 24
TOTAL number of itineraries 12 + 24 + 24 = 60
Answer: D
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To visit at least 2 of the 4 given European cities is to visit 2, 3, or all 4 cities. Since each itinerary must be made up of a different sequence of cities, order is important, and thus we have a permutation problem. The permutation formula is nPr = n! / (n - r)!. Let's determine the number of ways she can visit 2, 3, or all 4 cities.AAPL wrote:Veritas Prep
Anna has to visit at least 2 European cities on her vacation trip. If she can visit only London, Paris, Rome, or Madrid, how many different itineraries, defined as the sequence of visited cities, can Anna create?
A. 12
B. 36
C. 48
D. 60
E. 72
OA D
The number of ways she can visit exactly 2 cities is 4P2 = 4! / 2! = 4 x 3 = 12.
The number of ways she can visit exactly 3 cities is 4P3 = 4! / 1! = 4 x 3 x 2 = 24.
The number of ways she can visit all 4 cities is 4P4 = 4! / 0! = 4 x 3 x 2 x 1 = 24. (Recall that 0! = 1.)
Thus, the total number of ways she can visit at least 2 of the 4 cities is 12 + 24 + 24 = 60.
Answer: D
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