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Round Table ( Permutation and Combination)

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by GMATGuruNY » Thu Jun 28, 2012 7:42 pm
abcd1111 wrote:In how many ways 8 people can sit together around a round table if A and B always sit together .

Can anyone tell how to solve this type of question . I got Stuck !!

Thanks a lot in advance !!

Since AB must sit together, seat them at the table first.
Now count the number of ways that the 6 other people can be arranged around AB:
6! = 720.
Since AB can be reversed to BA, multiply by 2:
2*720 = 1440.
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by eagleeye » Thu Jun 28, 2012 7:47 pm
abcd1111 wrote:In how many ways 8 people can sit together around a round table if A and B always sit together .

Can anyone tell how to solve this type of question . I got Stuck !!

Thanks a lot in advance !!
Hi abcd1111:

Circular arrangements require that we fix one entity and arrange the rest.

For a simple case, let the 8 people sit without restriction. So we fix one person as the peg, and arrange the rest around that person. So we have 8-1 = 7 people left.

Then number of arrangements = (8-1)! =7!

Now in this particular case, consider the man and the woman as a single entity. Then we will do the arrangement. The two people can be arranged among themselves in 2!=2 ways. Let's fix those 2 as the single entity and arrange the rest (6 people who are left) around them.

In this case, the number of ways = 2!*6! = 2*720 = 1440.

I am going beyond the call of this question for clarity. Let's say if they asked for 5 people who must always sit together.

In that case, the 5 people who must sit together become the single entity that we will fix. The number of ways we can arrange these 5 people among themselves = 5!.
The number of ways in which we can arrange the remaining 3 = 3!.
Hence the total number of ways = 5!*3! = 120*6 = 720.

So, key takeaways are:
a. Fix one entity.
b. Find the number of ways of arranging that entity internally.
c. Find the number of ways of arranging the rest of the entities.
d. Multiply the two together.


Let me know if this helps :)
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