circular Permutation

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circular Permutation

by Manjareev » Wed Sep 26, 2012 3:10 am
3 ladies and 3 gents can be seated at a round table so that any two and only two of the ladies sit together. the number of ways is

a. 70

b. 27

c. 72

d. 48

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by GMATGuruNY » Wed Sep 26, 2012 4:40 am
Manjareev wrote:3 ladies and 3 gents can be seated at a round table so that any two and only two of the ladies sit together. the number of ways is

a. 70

b. 27

c. 72

d. 48
When people are seated at a circular table, where the first person sits is IRRELEVANT.
We need to count only the number of ways to arrange the remaining people RELATIVE to the first person seated.
Let the 3 women be A, B and C.

Case 1: A and B in adjacent seats
Once A is seated, the number of options for B = 2. (To the right or left of A.).
This AB block must be surrounded by men, so that 3 women are not in adjacent seats.
Number of options for the seat on the OTHER SIDE of A = 3. (Any of the 3 men.)
Number of options for the seat on the OTHER SIDE of B = 2. (Any of the 2 remaining men.)
Number of ways to arrange the 2 remaining people = 2! = 2.
To combine these options, we multiply:
2*3*2*2 = 24.

Remaining cases:
Since the same reasoning will apply to A and C in adjacent seats and to B and C in adjacent seats -- yielding 3 options for the two women in adjacent seats -- the result above must be multiplied by 3:
3*24 = 72.

The correct answer is C.
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by das.ashmita » Wed Sep 26, 2012 11:00 pm
Image

HI Manjareev

Here is my approach:

let us fix 2 ladies at position 1 & 2
The 3rd lady can occupy position 4 or 5 --> 2
Further, the 3 ladies can occupy the 3 positions in 3! ways --> 6
The men can occupy the remaining 3 positions in 3! ways --> 6


Therefore, total ways = 2*6*6 = 72

Ans C

Hope it helps :)

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by ashg84 » Thu Sep 27, 2012 6:21 am
Hi,

I got 72 as answer

My approach is as follows

In circular permutation no. of arrangement of n object is (n-1)!
So in this case
1. no. of ways of selecting two women out of 3, who will sit together will be 3c2 - 3
2. No. of ways if two women are together is (6 - 1- 1)! = 24
3. No. of ways of all three women sitting together (6-1-2)! = 6 & they came at 2 position before and after = total 6*2 = 12
So 2. - 3. = 24-12= 12
4. No. of ways two women can be arranged = 2! = 2

So total number of ways = 3*12*2 = 72

Please correct if i am wrong

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by Manjareev » Thu Sep 27, 2012 8:15 am
thank you so much GMATGURU n Ashmita.

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by eski » Sat Sep 29, 2012 3:41 am
1. No of was 2 ladies can be arranged = 2
2. No of was 4 men can be arranges is 3!
3. No of was all women (2 together) and 2 other , total 2 set elements is 3!

= 3!*3!*2 - 72