After a business trip to London, Michele has enough time to visit thre

Question

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

A. 36
B. 48
C. 72
D. 96
E. 120

abhimahna · Accepted Answer

Number of ways of selecting 3 cities = 6C3=20.
Lets say She decides to visit both Oslo and Stockholm and one other city, then number of ways of selecting the other city would be 4C1=4.

Thus, Number of ways of selecting three cities such that both Oslo and Stockhom are not chosen together = 20-4=16.

Since the question asks for the Sequence of the three cities, we will arrange the three cities to get all the possible outcomes.

Thus, Total different sequences will be 16 * 3! = 16*6 =96. Hence, the answer will be D.

Kurtosis · Answer

Hi  pallaviisinha ,

I will be able to help you with my explanation for your comment.

The question asks for the number of sequences. Ex: Paris, Barcelona and Rome is one sequence. Paris, Rome and Barcelona is a different sequence. So the question tests you on permutations.

Without restriction, the first city to be visited can be chosen from any of the 6 given cities. The next city can be chosen from the remaining 5 and the third city can be chosen from the remaining 4.

Total number of sequences without restriction = 6 * 5 * 4 = 120

Number of ways the cities can be visited when both Stockholm and Oslo are visited in a single trip --> Assume there are 3 slots. Oslo can be visited in any of the 3 slots, Stockholm can be visited in any of the 2 remaining slots and the remaining slot can be chosen among 4 different cities --> Number of sequences = 3 * 2 * 4 = 24

Number of sequences the cities can be visited such that Oslo and Stockholm are not visited in a single trip = Total - Number of sequences the cities can be visited such that Stockholm and Oslo are visited in a single trip = 120 - 24 = 96.

Hope it helps.

Divyadisha · Answer

Sequence of cities she can visit without any restrictions= 6*5*4= 120

Sequence of cities she can visit if she visits both Oslo and Stockholm on the same trip= 3*2 *1 *4 (after choosing Oslo and Stockholm, she is left with 4 cities to choose from)= 24

Sequence of cities to choose with restrictions= 120-24= 96

D is the answer

pallaviisinha · Answer

Hi Divyadisha,

Pls explain how did you arrive at the following sentence:

Sequence of cities she can visit if she visits both Oslo and Stockholm on the same trip= 3*2 *1 *4 (after choosing Oslo and Stockholm, she is left with 4 cities to choose from)= 24

Thanks, 
Pallavi

abhijeetkaushik · Answer

Why have you conveniently decided to place Oslo, and Stockholm first? What if I were to arrange in this way :

_ _ _
So I have three slots, the "other" city can be visited in any of those slots. Let's say I choose the first slot, then I have 4 options. With the second slot, I have only 2 options (either Oslo/Stockholm), and with the third I have only one. That gives a total of 8 ways.

Kurtosis · Answer

You are restricting the 'others' to only first slot. You can initially place in any of the 3 slots. So it's 3 * 4 = 12.

The second slot will have 2 choices.

The third slot will have only one choice.

12 * 2 * 1 = 24

Sent from my ONE A2003 using Tapatalk

Scuven · Answer

You can also solve this in another way other than reversal method (Total - arranged combinations of O and S together).

You can use FCF like this: you have to fill three places:

- 6 cities for 1st spot
- 4 cities for second spot 
- 2 cities for third spot

You jump 5 and 3 because you exclude O and S. Thus you obtain 6*4*2= 48 ways of arranging and combining without O and S. Then you consider that the couple O and S can have two arrangements 1) O-S 2)S-O. Thus you multiply 48 *2!=96 -> D

AakankshaArora · Answer

If we divide two sequences of trips 1) with Oslo and 2) with S 
then by filling space method we will have O*3*4 and S*3*4 = 12+12= 24 
And since the sequence matters we can multiply by 3!
can someone please tell me where i go wrong.

Fdambro294 · Answer

We are asked for the different possible sequences, the is means that the different arrangements of the 3 trips chosen will count as different combinations (in other words, “order matters”)

(Total Outcomes possible) - (No. of Unfavorable Outcomes in which O and S are included) = no. of favorable outcomes

(1st)we can choose 3 of the 6 cities in:

6! / (3! 3!) = 20 different ways

Then, for each different way, we can arrange the 3 cities chosen in ——-> 3! = 6 different sequences.

(20) (6) = 120 total outcomes with no constraint

(2nd)to find the number of Unfavorable Outcomes, we can make sure that for each outcome O and S are already picked. This leave 4 other cities and only 1 city left to be picked ——-> 4! / (1! 3!) = 4 ways

And, for each of these ways, we can arrange the 3 cities in ——-> 3! = 6 different sequences

(4) (6) = 24 unfavorable outcomes

Answer:

120 - 24 =

96

Posted from my mobile device

KarishmaB · Answer

Of the 6 cities, 3 can be selected in 6C3 ways = 20 ways.
Of these, some selections are not acceptable i.e. those in which both Oslo and Stockholm are selected. This selection can be made in 4C1 ways (select Oslo, Stockholm and 1 of the leftover 4 cities)

No of acceptable selections = 20 - 4 = 16

The 3 cities can be arranged in 3! ways to give us total 16 * 3! = 96 sequences

Answer (D)

HaSmamit · Answer

Hi  KarishmaB ,
I know it's an old thread but would appreciate your input.
Let's say I don't approach this from a "1-bad choice" approach, but try to sum the good choices.

Why is what I'm doing wrong?

She could either choose 3 destinations that do not include the forbidden two at all, and there should be 4*3*2*3! options.
Or, she could choose to visit either Oslo or Stockholm - in which case there'd be 1*4*3*3! options.

Adding those up - 4*3*2*3!+2*1*4*3*3! results in a big number which is very far from the answer. Could you please explain to me what I'm doing wrong?

KarishmaB · Answer

You are arranging them twice. In the first case, you are arranging 4 cities (by using fundamental counting principle 4*3*2*1) and then you are arranging 3 cities (using 3!) and multiplying them. You need to SELECT 3 cities out of 4 and that you can do in 4C3 ways which gives us only 4. Now you have to arrange these 3 cities and that can be done in 3! ways. So you get 4 * 3! = 24 Next you select Oslo and select 2 more from the other 4 in 4C2 ways. You then arrange all 3 selected cities in 3! ways. You get 4C2 *3! = 36 Same for Stockholm so another 36 Total = 24 + 36 + 36 = 96 Check out these two videos on my YouTube channel. They discuss the basics of fundamental counting principle and combinations formula: Video on Permutations: https://youtu.be/LFnLKx06EMU Video on Combinations: https://youtu.be/tUPJhcUxllQ

Capablanca · Answer

In case someone wonders, this would be the direct approach:

Number of ways to arrange without Oslo and Stockholm:  There are 6-2 options so 4 x 3 x 2 = 24

Number of ways to arrange including Oslo:  1 x 4 x 3 = 12 and multiply by 3

To illustrate:

If Oslo is first (O _ _), we have 4 choices for the second city (excluding Stockholm) and 3 choices for the third. That's 1 x 4 x 3 = 12.

If Oslo is second (_ O _), we have 4 choices for the first city and 3 choices for the third. That's 4 x 3 = 12.

If Oslo is third (_ _ O), we have 4 choices for the first city and 3 choices for the second. That's 4 x 3 = 12

So, the total number of ways to arrange the trip, including Oslo, is 3 x 12 = 36.

Finally, we use the same approach to calculate Stockholm and the answer is 36+36+24 = 96

After a business trip to London, Michele has enough time to visit thre

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

After a business trip to London, Michele has enough time to visit thre

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards