GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 23 Jan 2020, 16:30

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# A man wants to visit at least two of the four cities A, B, C

Author Message
TAGS:

### Hide Tags

Intern
Joined: 03 Jan 2014
Posts: 48
Concentration: General Management
Schools: Schulich '16
GMAT 1: 690 Q49 V34
A man wants to visit at least two of the four cities A, B, C  [#permalink]

### Show Tags

Updated on: 02 Feb 2014, 02:35
14
00:00

Difficulty:

85% (hard)

Question Stats:

46% (01:32) correct 54% (01:26) wrong based on 225 sessions

### HideShow timer Statistics

A man wants to visit at least two of the four cities A, B, C and D. How many travel itineraries can he make? All cities are connected to one another.

(A) 24
(B) 6
(C) 60
(D) 12
(E) None of the above

Originally posted by vimal096 on 01 Feb 2014, 22:33.
Last edited by Bunuel on 02 Feb 2014, 02:35, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
Math Expert
Joined: 02 Sep 2009
Posts: 60627
Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

### Show Tags

02 Feb 2014, 02:40
5
5
vimal096 wrote:
A man wants to visit at least two of the four cities A, B, C and D. How many travel itineraries can he make? All cities are connected to one another.

(A) 24
(B) 6
(C) 60
(D) 12
(E) None of the above

The # of ways to visit 2 cities when order matters = $$P^2_4=\frac{4!}{(4-2)!}=12$$: AB, BA, AC, CA, AD, DA, BC, CB, BD, DB. CD, DC.

The # of ways to visit 3 cities when order matters = $$P^3_4=\frac{4!}{(4-3)!}=24$$: ABC, ACB, BAC, BCA, CAB, CBA, ...

The # of ways to visit 4 cities when order matters = $$P^4_4=\frac{4!}{(4-4)!}=24$$: ABCD, ABDC, ADCD, ...

12 + 24 + 24 = 60.

_________________
##### General Discussion
Retired Moderator
Joined: 17 Sep 2013
Posts: 317
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE: Analyst (Consulting)
Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

### Show Tags

22 Apr 2014, 10:43
2
We can also say:

4C2*2! + 4C3*3! + 4C4*4!

I am quite confused with this P n C terminologies..on where to use...so I go intutively..i.e First select the elements and then shuffle them with factorial (If required)...Veritas has an excellent guide on combinatorics..which combined with Bunuels Q bank is enough for 700+ Q on this topic
Math Expert
Joined: 02 Sep 2009
Posts: 60627
Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

### Show Tags

22 Apr 2014, 11:01
1
JusTLucK04 wrote:
We can also say:

4C2*2! + 4C3*3! + 4C4*4!

I am quite confused with this P n C terminologies..on where to use...so I go intutively..i.e First select the elements and then shuffle them with factorial (If required)...Veritas has an excellent guide on combinatorics..which combined with Bunuels Q bank is enough for 700+ Q on this topic

This is correct. $$C^2_4*2!$$ is the same as $$P^2_4$$.
_________________
Manager
Joined: 26 Apr 2015
Posts: 123
Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

### Show Tags

20 Aug 2015, 23:04
Bunuel wrote:
vimal096 wrote:
A man wants to visit at least two of the four cities A, B, C and D. How many travel itineraries can he make? All cities are connected to one another.

(A) 24
(B) 6
(C) 60
(D) 12
(E) None of the above

The # of ways to visit 2 cities when order matters = $$P^2_4=\frac{4!}{(4-2)!}=12$$: AB, BA, AC, CA, AD, DA, BC, CB, BD, DB. CD, DC.

The # of ways to visit 3 cities when order matters = $$P^3_4=\frac{4!}{(4-3)!}=24$$: ABC, ACB, BAC, BCA, CAB, CBA, ...

The # of ways to visit 4 cities when order matters = $$P^4_4=\frac{4!}{(4-4)!}=24$$: ABCD, ABDC, ADCD, ...

12 + 24 + 24 = 60.

Why not 4C2x4C3x4C4?
Manager
Joined: 10 Jun 2015
Posts: 110
Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

### Show Tags

20 Aug 2015, 23:42
vimal096 wrote:
A man wants to visit at least two of the four cities A, B, C and D. How many travel itineraries can he make? All cities are connected to one another.

(A) 24
(B) 6
(C) 60
(D) 12
(E) None of the above

At least 2 means 2 or more
2 cities can be selected from 4 in 4C2 ways and these two cities, say, A and B have two distinct itineraries AB and BA
That can be done in 4C2*2! or 4P2 ways=12
Similarly, 3 cities in 4C3*3! or 4P3 ways=24; 4 cities in 4C4*4! or 4P4 ways=24
Total 12+24+24=60 ways.
The correct option is C
Intern
Joined: 28 Dec 2015
Posts: 37
Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

### Show Tags

21 Jul 2016, 03:59
SonofAnarchy wrote:
Bunuel wrote:
vimal096 wrote:
A man wants to visit at least two of the four cities A, B, C and D. How many travel itineraries can he make? All cities are connected to one another.

(A) 24
(B) 6
(C) 60
(D) 12
(E) None of the above

The # of ways to visit 2 cities when order matters = $$P^2_4=\frac{4!}{(4-2)!}=12$$: AB, BA, AC, CA, AD, DA, BC, CB, BD, DB. CD, DC.

The # of ways to visit 3 cities when order matters = $$P^3_4=\frac{4!}{(4-3)!}=24$$: ABC, ACB, BAC, BCA, CAB, CBA, ...

The # of ways to visit 4 cities when order matters = $$P^4_4=\frac{4!}{(4-4)!}=24$$: ABCD, ABDC, ADCD, ...

12 + 24 + 24 = 60.

Why not 4C2x4C3x4C4?

Travel itinerary means you can start your travel from any city-say you start from B and then moved to A and other way round you can start from A and move to B-so you have two different itineraries.Itineraries literally means planned routes and so you have two different routes as I explained.
Intern
Joined: 15 Dec 2016
Posts: 2
Location: India
Concentration: Finance, Entrepreneurship
GMAT 1: 400 Q43 V6
GPA: 2.5
A man wants to visit at least two of the four cities A, B, C  [#permalink]

### Show Tags

10 Jan 2017, 05:17
In this question ALL THE CITIES ARE CONNECTED TO ONE ANOTHER, .How you consider this piece of information.I am not getting this part. I am supposed to select at least two cities. could you please just give me a brief idea. how to solve this type of questions.
Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1344
Location: Malaysia
A man wants to visit at least two of the four cities A, B, C  [#permalink]

### Show Tags

10 Jan 2017, 05:37
JusTLucK04 wrote:
We can also say:

4C2*2! + 4C3*3! + 4C4*4!

I am quite confused with this P n C terminologies..on where to use...so I go intutively..i.e First select the elements and then shuffle them with factorial (If required)...Veritas has an excellent guide on combinatorics..which combined with Bunuels Q bank is enough for 700+ Q on this topic

Bunuel Why we need 4C2*2!+ 4C3*3! + 4C4*4! as highlighted?
_________________
"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Intern
Joined: 08 Oct 2017
Posts: 30
Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

### Show Tags

10 Jul 2018, 09:06
hazelnut
Ways to choose cities x ways to travel
At least 2 cities —> 6 ways (2 cities chose out 4) x 2! (Ways to travel to cities) = 12
3 cities —> 4 x 3! =24
4 cities —> 1 x 4! = 24
Total = 60 (C)
Non-Human User
Joined: 09 Sep 2013
Posts: 14003
Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

### Show Tags

10 Dec 2019, 03:21
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: A man wants to visit at least two of the four cities A, B, C   [#permalink] 10 Dec 2019, 03:21
Display posts from previous: Sort by