carlos.lara.7 wrote:After a business trip to London, Michele has enough time to visit three European cities before returning home. If she has narrowed her list to six cities that she'd like to visit - Paris, Barcelona, Rome, Munich, Oslo, and Stockholm - but does not want to visit both Oslo and Stockholm on the same trip, how many different sequences of three cities does she have to choose from?

a)36

b)48

c)72

d)96

e)120

Good arrangements = all possible arrangements - bad arrangements.

All possible arrangements:
Number of options for the first visited city = 6. (Any of the 6 cities.)
Number of options for the second visited city = 5. (Any of the 5 remaining cities.)
Number of options for the third visited city = 4. (Any of the 4 remaining cities.)
To combine the options in blue, we multiply:
6*5*4 = 120.

Bad arrangements:
A bad arrangement includes both Oslo and Stockholm.
Number of options for the third city to be combined with Oslo and Stockholm = 4. (Any of the 4 other cities.)
Number of ways to arrange the 3 cities = 3! = 6.
To combine the options in blue, we multiply:
4*6 = 24.

Good arrangements:
All possible arrangements - bad arrangements = 120-24 = 96.

The correct answer is D.

An alternative way to do this one is to add all possible arrangements.

Arrangements Without Oslo Or Stockholm:

4 Cities - Paris, Barcelona, Rome and Munich

4P3 = 4 x 3 x 2 = 24

Arrangements Including Oslo But Not Stockholm:

Oslo with 2 Of the 4 Others

Oslo with 4C2 = 6 Different Combinations Of Cities

Each of the 6 combinations of 3 cities can be arranged in 3! = 6 ways.

6 x 6 = 36 Different Arrangements Of Oslo With 2 Other Cities

Arrangements Including Stockholm But Not Oslo:

This works the same as the arrangements including Oslo but not Stockholm.

36 Different Arrangements Of Stockholm With 2 Other Cities

Total:

24 + 36 + 36 = 96 Different Trips

The correct answer is D.

Another approach:

Michele doesn't visit either Stockholm or Oslo: 4 options * 3 options * 2 options => 24 options.

Michele visits Stockholm, but not Oslo: 1 option * 4 options * 3 options * 3 (since Stockholm could be first, second, or third) => 36 options

Michele visits Oslo, but not Stockholm: same as the case above, so => 36 options.

We're left with 24 + 36 + 36 => 96 options.