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

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

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New post Updated on: 02 Feb 2014, 02:35
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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.
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Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

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New post 02 Feb 2014, 02:40
4
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.

Answer: C.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 2, 3, 5, 7 and 10. Thank you.
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Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

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New post 22 Apr 2014, 10:43
1
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
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Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

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New post 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 \(P^2_4\).
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Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

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New post 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.

Answer: C.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 2, 3, 5, 7 and 10. Thank you.



Why not 4C2x4C3x4C4?
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Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

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New post 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
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Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

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New post 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.

Answer: C.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 2, 3, 5, 7 and 10. Thank you.



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

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New post 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.
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A man wants to visit at least two of the four cities A, B, C  [#permalink]

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New post 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?
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Re: A man wants to visit at least two of the four cities A, B, C  [#permalink]

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New post 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)
Re: A man wants to visit at least two of the four cities A, B, C &nbs [#permalink] 10 Jul 2018, 09:06
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