3 ladies and 3 gents can be seated at a round table so that any two an...
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.
View all questions of this test
3 ladies and 3 gents can be seated at a round table so that any two an...
Out of the 3 ladies, if 2 are to sit together they can be seated in 3P2 = 6 ways.
Now, in the seats adjacent to the ladies only 2 gents can be seated because only 2 ladies are supposed to sit together.
Out of 3 gents, if 2 are to sit together they can be seated in 3P2 = 6 ways.
Now, the remaining two people can be seated in the remaining two seats in 2P2 = 2 ways.
The number of ways in which the 3 ladies and 3 gents can be seated at a round table so that any 2 and only 2 of the ladies sit together are
= 3P2 × 3P2 × 2P2
= 6 × 6 × 2
= 72.
3 ladies and 3 gents can be seated at a round table so that any two an...
Given: 3 ladies and 3 gents can be seated at a round table so that any two and only two of the ladies sit together.
To find: The number of ways the people can be seated around the table.
Solution:
Arrangement of 6 people around a circular table can be done in (6-1)! = 5! ways.
Let's consider the ladies as L1, L2, L3 and gents as G1, G2, G3.
Case 1: L1, L2, L3 sit together
In this case, we can consider L1, L2, L3 as a single unit, and arrange them and the gents around the table in (4-1)! = 3! ways.
But within the unit L1, L2, L3, the ladies can be arranged in 3! ways.
Therefore, the total number of arrangements in this case = 3! x 3! = 36
Case 2: Exactly two ladies sit together
In this case, we can choose any two ladies out of the three in 3C2 ways.
Consider the chosen ladies as a single unit, and arrange them and the gents around the table in (4-1)! = 3! ways.
Within the unit of chosen ladies, the ladies can be arranged in 2! ways.
The remaining lady can be seated at any of the 4 gaps between the units or at the beginning/end of the unit in 4 ways.
Therefore, the total number of arrangements in this case = 3C2 x 3! x 2! x 4 = 72
Case 3: No two ladies sit together
In this case, we need to arrange the gents and the ladies in an alternate manner around the table.
The number of ways to arrange the gents = 3!
The number of ways to arrange the ladies = 3!
Therefore, the total number of arrangements in this case = 3! x 3! = 36
Total number of arrangements = Sum of arrangements in all cases
= 36 + 72 + 36
= 144
But, the table is circular, so each arrangement can be rotated in 6 ways.
Therefore, the number of distinct arrangements = 144/6 = 24
Hence, the correct option is (C) 72.
To make sure you are not studying endlessly, EduRev has designed CA Foundation study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in CA Foundation.