Jill, who lives in City C, plans to visit 3 different cities, M, L, an

She can visit the cities in any order. Number of ways of arranging the three cities is 3! = 6 ways. Hence B.
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3! = 6

M L S C
M S L C
L M S C
L S M C
S M L C
S L M C

Question is asking for order --> means permutation is needed

Hence, 3x2x1 = 6

No of ways Jill can visit the cities and returns is 3*2*1 => 6 ways...

Hence answer will be (B) 6
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hi
abhi , although i got it right but can you please correct me where i am wrong , see it is mentioned that order does not matter here then the answer would be like
3!/2! that =3 and now she is coming back then same for this =3+3= 6 ways

please correct me where i am wrong , because you all guys are doing only 3! , which seems that order does matter here

thanks

Hi nks2611 ,

No, your reasoning is not correct. We are given she visits each of the city only once and then comes directly back to city C. So, your point "she is coming back then same for this =3" is incorrect as there will be only one way to reach each city.

So, we have the form of L,M,N and Last C.

No. of ways to arrange L,M and N when order doesn't matter = 3! = 6 ways.

Notice that C will remain as it is.

If I talk in your terms, "order does not matter here then the answer would be like 3!/2! that =3 ", it should not be 3P2 but it should be 3P1 as she is selecting one city out of 3 to go to and not 2 cities.

I hope that makes sense.

If not, reach out with your queries.
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Thank very much abhi i got my mistake, here she can travel from c to any other citu for once so the order can be difffernt for this so total ways should be 6 ways

Sent from my vivo 1601 using GMAT Club Forum mobile app

The question says---Jill, who lives in City C, plans to visit 3 different cities, M, L, and S. She plans to visit each city exactly once and return to City C after the 3 visits.

Thus it is very clear that jill comes back to C after 3 city visits.We need to count C in one of these 3 city visits.

C
- - - = considering M,L and S in first 2 places(3P2) and C in its last place(1!)--->total number of arrangements=3P2*1!=6

Hope this approach is right.

--Sujith

6 orderings

C MLS C
C MSL C
C LSM C
C LMS C
C SLM C
C SML C

Since we care about the order in which Jill visits the cities, this is a permutations question.

Here's an explanation of this question on YouTube:

https://www.youtube.com/watch?v=U1V029s ... rjxKZWMPuo

Note that if we didn't care about the order - if this were a combinations question - the answer would just be 1!

Hello,

Can someone explain this question using the permutation formula P(n,r)= n¡/(n-r)¡. thanks a lot.

I thought of the journey as a circular one. Therefore (n-1)! is (4-1)! = 6

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