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Expert Help Required on Counting Problem

by knight247 » Tue Jul 30, 2013 9:08 am
There are 5 people - A, B, C, D and E. They have to sit around a circular table with 5 chairs such that A can sit neither next to D nor next to E. How many such distinct arrangements are possible?


I can solve the problem by the regular method, where one seats : A in 1 way; B in 2 ways (on either side of A); C in 1 way (on the other side of A, as compared to B); D in any one of 2 ways; E in one way.

1*2*2*1 = 4

However, I'm unable to solve it via the method :

Desired Outcomes = Total possible outcomes - Undesirable outcomes

Hoping to get a detailed explanation on how to solve via this method, preferably by an expert. Many thanks in advance.
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by Matt@VeritasPrep » Tue Jul 30, 2013 9:23 am
First off, this method of solving this question is ill-advised: it makes the problem MUCH more complicated. A better way of solving is to say that A must be between B and C (which can only happen in two ways: BAC or CAB) and that D and E can only be seated in the other two seats between B and C in two ways (DE or ED), giving us 2 * 2 solutions.

However ...

Using the formula you're after (which has a fancy name, The Principle of Inclusion and Exclusion, and a friendlier nickname, PIE), we have

Total Arrangements - (# without A and D together) - (# without A and E together) + (# without A, D, and E together) = Valid Arrangements

Total = 4!

# without A and D together = 3*2*2
# without A and E together = 3*2*2
# without A, D, and E together = 2*2

4! - 12 - 12 + 4 = 4
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by vishugogo » Tue Jul 30, 2013 11:35 am
kindly explain without using this formula
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by Matt@VeritasPrep » Tue Jul 30, 2013 1:40 pm
Since A can't be seated next to D or E, A must be in between B and C. Thinking of the circle, this means either:

1) B is on A's left and C is on A's right
or
2) C is on A's left and B is on A's right

So there are two ways to arrange B and C.

D and E must be across the table from A, and there are two ways to arrange them:

1) D is on E's left
2) D is on E's right

Since there are 2 ways to arrange each group, there are 2 * 2 = 4 ways to arrange everyone relative to one another (which is all that matters in a generic circular arrangement).
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by GMATGuruNY » Tue Jul 30, 2013 2:02 pm
knight247 wrote:There are 5 people - A, B, C, D and E. They have to sit around a circular table with 5 chairs such that A can sit neither next to D nor next to E. How many such distinct arrangements are possible?
For circular arrangements, we count the number of ways to arrange the REMAINING people RELATIVE to the first person seated.

Once A is seated:
Number of options for the seat to A's left = 2. (Must be B or C.)
Number of options for the seat to A's right = 1. (Either B or C, whoever isn't seated to A's left.)
Number of options for the next seat = 2. (Either C or D.)
Number of options for the last seat = 1. (Only 1 person left.)
To combine these options, we multiply:
2*1*2*1 = 4.
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by vipulgoyal » Wed Jul 31, 2013 11:05 pm
Mitch/Matt,

Please explain whats wrong with this

just forget about A, ways to arrange B,C,D,E in circular arrangement = 3!
now only one place left for A which meets the condition therefore 3!x1 = 6
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by GMATGuruNY » Thu Aug 01, 2013 6:22 am
vipulgoyal wrote:Mitch/Matt,

Please explain whats wrong with this

just forget about A, ways to arrange B,C,D,E in circular arrangement = 3!
now only one place left for A which meets the condition therefore 3!x1 = 6
The 6 ways to arrange B, C, D and E in a circle include 2 orderings that are not allowed:
B - D - C - E
B - E - C - D
In these arrangements, there is no way to place A such that A will not be adjacent to either D or E.
Since these two orderings are not valid, the total number of viable arrangements = 6-2 = 4.
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