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permutations and combinations

Expert replies
Source: — Problem Solving |

by smanstar » Tue Sep 25, 2012 9:12 pm
Consider the 3 Americans and 2 Canadian as a single entity as they always are together.

So we have the following pattern ACI which can be arranged in (3!) ways = 6 ways.

Further the 3 Americans and 2 Canadians can be arranged in 3! and 2! ways among/between them.

So required no of arrangements = 6 * 6 * 2 = 72 ways
Last edited by smanstar on Wed Sep 26, 2012 7:40 am, edited 1 time in total.
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by smanstar » Tue Sep 25, 2012 9:26 pm
hey sorry for the 1st reply I misread the question as 2 Indians , it is rectified now
Hope it helps
Last edited by smanstar on Wed Sep 26, 2012 10:28 pm, edited 1 time in total.
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by Brent@GMATPrepNow » Wed Sep 26, 2012 6:56 am
srivathsan wrote:In how many ways can 3 canadians, 2 americans and an indian can be arranged in a row such that all 3 canadians and 2 americans are always together?
Hmm, there's some ambiguity here.
Are you saying that the 3 Canadians must always sit together, and the 2 Americans must always sit together? (e.g., C,C,C,I,A,A would be fine)
Or, are you saying the 5 people who are either Canadian or American must sit together. (e.g., C,A,A,C,C,I would be fine)

I'm assuming that it's the first one. That is, the 3 Canadians must always sit together, and the 2 Americans must always sit together? (e.g., C,C,C,I,A,A would be fine)

Take the task of seating everyone and break it into stages.
Stage 1: Arrange the 3 Canadians in a row.
We have a nice rule for this. We can arrange n unique objects in a row in n! ways.
So, we can arrange the 3 Canadians in 3! ways (6 ways).

Stage 2: Arrange the 2 Americans in a row.
We can complete this stage in 2! ways (2 ways).

Stage 3: Arrange the 1 Indian in a row.
We can complete this stage in (1 way).

Stage 4: Take the pre-arranged groups of Canadians, Americans and Indian, and arrange them in a row.
There are 3 groups, so we can arrange them in 3! ways (6 ways).

By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus seat all 6 people) in (6)(2)(1)(6) ways ([spoiler]= 72 ways[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775
Brent Hanneson - Creator of GMATPrepNow.com
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by Brent@GMATPrepNow » Sat Sep 29, 2012 9:18 am
In my opinion, the majority of counting questions can be solved using the Fundamental Counting Principle (FCP). If you're interested in practicing this technique, here are a few questions to try:

https://www.beatthegmat.com/permutation-t122873.html
https://www.beatthegmat.com/a-certain-ship-t116519.html
https://www.beatthegmat.com/permutation-t112874.html
https://www.beatthegmat.com/permutation- ... 92779.html
https://www.beatthegmat.com/combinations ... 15538.html
https://www.beatthegmat.com/3-rings-t92252.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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