Permutation or Combination?

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s.pharmaco: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Sun Jul 08, 2012 12:35 am

by s.pharmaco » Thu Dec 13, 2012 7:58 am

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

Can somebody please explain why we need to use permutation here and not combination?

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Source: Beat The GMAT — Problem Solving | Thu Dec 13, 2012 7:58 am

Anindya Madhudor: Senior | Next Rank: 100 Posts; Posts: 85; Joined: Mon Nov 12, 2012 11:12 amFollowed by: 2 members; Thanked: 8 times

by Anindya Madhudor » Thu Dec 13, 2012 8:27 am

Say the man wants to visit just 2 cities of the 4 cities. Then, you have the following 6 options:

AB AC AD BC BC CD

This 6 options can be obtained by using the combination formula 4C2.

But, in this case, itinerary means AB is not the same as BA, which means you can start with either city A or city B and have two different itinerary. Hence the problem becomes a permutation problem. So, you need to multiply 6 by 2 to obtain 12 as answer for ways you can form an itinerary to visit 2 cities out of 4 cities.

Hope this makes sense.

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s.pharmaco: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Sun Jul 08, 2012 12:35 am

by s.pharmaco » Thu Dec 13, 2012 9:05 am

Thanks for your reply. so in this question is "all cities are connected to one another" the cue to decide what we need to apply here (P or C)?

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Anindya Madhudor: Senior | Next Rank: 100 Posts; Posts: 85; Joined: Mon Nov 12, 2012 11:12 amFollowed by: 2 members; Thanked: 8 times

by Anindya Madhudor » Thu Dec 13, 2012 9:19 am

If the question asks for "how many ways the man can choose 2 or more cities for his itinerary", then it would be a combination problem. The key here is that starting from city A is not the same as starting from city B. So, AB and BA are different itineraries.

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s.pharmaco: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Sun Jul 08, 2012 12:35 am

by s.pharmaco » Thu Dec 13, 2012 10:07 am

okies cool now understood.. Thanks...

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puneetkhurana2000: Master | Next Rank: 500 Posts; Posts: 131; Joined: Wed Nov 14, 2012 2:01 pmFollowed by: 2 members; Thanked: 39 times

by puneetkhurana2000 » Thu Dec 13, 2012 12:22 pm

Good work Anindya Madhudor!!!

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Anindya Madhudor: Senior | Next Rank: 100 Posts; Posts: 85; Joined: Mon Nov 12, 2012 11:12 amFollowed by: 2 members; Thanked: 8 times

by Anindya Madhudor » Thu Dec 13, 2012 1:19 pm

Thank you, Puneet!

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The Iceman: Master | Next Rank: 500 Posts; Posts: 194; Joined: Mon Oct 15, 2012 7:14 pmFollowed by: 6 members; Location: India; Thanked: 47 times

by The Iceman » Fri Dec 14, 2012 4:49 am

s.pharmaco wrote:A man wants to visit at least two of the four cities A, B, C and D. how many travel itineries can he make? all cities are connected to one another.

Can somebody please explain why we need to use permutation here and not combination?

It's basically a problem on arrangements because order matters here.

First you can arrange all 4 cities. Then select three cities out of four and arrange those three cities. Lastly, select two out of the four and arrange those two cities.

4!+(4C3)*3!+4C2*2! = 60

You could also see this as 4P4 + 4P3 + 4P2, but it is easier to follow the convention using combination "C", given you are bale to apply the logic correctly.

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mneeti: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Sun Jul 08, 2012 1:09 am

by mneeti » Mon Jul 22, 2013 5:19 am

The Iceman wrote:
s.pharmaco wrote:A man wants to visit at least two of the four cities A, B, C and D. how many travel itineries can he make? all cities are connected to one another.

Can somebody please explain why we need to use permutation here and not combination?
It's basically a problem on arrangements because order matters here.

First you can arrange all 4 cities. Then select three cities out of four and arrange those three cities. Lastly, select two out of the four and arrange those two cities.

4!+(4C3)*3!+4C2*2! = 60

You could also see this as 4P4 + 4P3 + 4P2, but it is easier to follow the convention using combination "C", given you are bale to apply the logic correctly.

I tried a different approach to solve this question, though its wrong but I want to understand why my approach is wrong here. Please help on that!

There are 4 cites,so for each city we have 2 options either 'Yes' or 'No' making the no. of ways available = 2x2x2x2 = 16

Now, we have to make itineraries picking at least 2 cities. This can be done by eliminating 1) Where we have 'No' in all the cities, there is only one way it can happen=1 2)Where only one city is not picked, this can happen in four ways [YYYN] [YYNY] [YNYY] [NYYY] =4.

Therefore, the no. of ways itineraries can be made picking at least two cities = 16-1-4=11.

What is wrong with my approach?

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GMAT/MBA Expert

[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members; Location: Palo Alto, CA; Thanked: 2867 times; GMAT Score:800

by [email protected] » Mon Jul 22, 2013 10:11 am

Hi mneeti,

Your approach doesn't account for the ORDER of the stops on the itinerary.

Going to City A first and City B second is NOT THE SAME as going to City B first and City A second.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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vipulgoyal: Master | Next Rank: 500 Posts; Posts: 468; Joined: Mon Jul 25, 2011 10:20 pmFollowed by: 4 members; Thanked: 29 times

by vipulgoyal » Tue Jul 23, 2013 2:29 am

My take 40,
the qustion asks how many travel itineries can he make if he wants to visit at least 2 cities
it means differant itineries for visiting 2 cities + differant itineries for visiting 3 cities

for 2 cities (since all cities are connected to one another)
A-B-C "A to B to C"
A-C-D
A-D-B
A-B-C
Like starting with A we have 4 itineries same for B,C and D
4+4+4+4 = 16
Smae as for 3 cities

fix A in first position, rest B,C and D can be rearranges in 6 ways
same as for B,C, and D

6+6+6+6 = 24

24 + 16 = 40 Inineries are possible

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