Question

If 3 sisters and 8 brothers are together playing a game, then the number of ways in which all the sisters and brothers are to be seated around a circle such that all the three sisters are not seated together is

• A. \( 8! \times 504 \)
• B. \( 11! \times 8 \)
• C. \( 7! \times 210 \)
• D. \( 8! \times 84 \)

Hint:

For circular permutation problems involving restrictions like "not seated together", it's crucial to understand the exact meaning of the restriction. \begin{itemize} \item \textbf{If "all elements are not together" (e.g., all 3 sisters are not together):} Calculate the total arrangements and subtract arrangements where all those specific elements are together. This is the approach used in the solution. \item \textbf{If "no two elements are together" (e.g., no two sisters sit together):} Use the gap method. First, arrange the other elements, then place the restricted elements in the gaps created. \end{itemize} In this problem, "all the three sisters are not seated together" implies that the block of 3 sisters as a unit is broken.

Answer 1

The Correct Option is D

Let S be the number of sisters, so \( S = 3 \). Let B be the number of brothers, so \( B = 8 \). Total number of people \( N = S + B = 3 + 8 = 11 \). The total number of ways to arrange \( N \) distinct people around a circle is \( (N-1)! \). So, the total number of ways to arrange all 11 people around a circle without any restrictions is \( (11-1)! = 10! \). Now, we need to find the number of ways such that all three sisters are NOT seated together. This can be found by subtracting the number of ways where all three sisters ARE seated together from the total number of arrangements. Case: All three sisters are seated together. Treat the 3 sisters as a single block or unit. Now we have 1 unit (of sisters) + 8 brothers = 9 units. Arrange these 9 units around a circle. The number of ways to do this is \( (9-1)! = 8! \). Within the block of 3 sisters, they can arrange themselves in \( 3! \) ways. So, the number of ways where all three sisters are seated together is \( 8! \times 3! \). Number of ways where all three sisters are NOT seated together = (Total arrangements) - (Arrangements where all three sisters are together) \( = 10! - (8! \times 3!) \) \( = 10 \times 9 \times 8! - 8! \times 3! \) \( = 8! (10 \times 9 - 3!) \) \( = 8! (90 - 6) \) \( = 8! (84) \) Let's check the options: % Option (A) \( 8! \times 504 \) % Option (B) \( 11! \times 8 \) % Option (C) \( 7! \times 210 \) % Option (D) \( 8! \times 84 \) Our calculated answer matches option (D). Alternative Method: Gap Method First, arrange the 8 brothers around a circle. The number of ways to do this is \( (8-1)! = 7! \). When 8 brothers are arranged around a circle, there are 8 gaps between them. For the 3 sisters to not be seated together, they must be placed in these gaps, with at most one sister per gap. We have 8 gaps and we need to choose 3 of them for the sisters, which can be done in \( P(8,3) \) ways (permutation because the sisters are distinct). \( P(8,3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336 \). So, the number of ways is \( 7! \times P(8,3) = 7! \times 336 \). Let's convert this to the format of the options: \( 7! \times 336 = 7! \times (4 \times 84) = 4 \times 7! \times 84 = (4 \times 7!) \times 84 \). This doesn't seem to directly match. Let's convert \( 7! \times 336 \) to \( 8! \times 84 \): \( 7! \times 336 = 7! \times (8 \times 42) \). This doesn't seem to simplify easily to the target format. Let's re-evaluate \( 7! \times P(8,3) \): \( 7! \times 8 \times 7 \times 6 = (8 \times 7!) \times (7 \times 6) = 8! \times 42 \). This result is \( 8! \times 42 \), which is not \( 8! \times 84 \). Let's recheck the "all three sisters are not seated together" interpretation. "All three sisters are not seated together" typically means no two sisters are adjacent, OR it can mean that the block of three sisters is not together. The phrasing "all the three sisters are not seated together" usually implies that the block of 3 sisters is broken, but not necessarily that no two sisters are adjacent. However, in competitive exams, this phrase often means "no two sisters are together" if the context implies maximum separation. If it means "the block of 3 sisters is not formed", then the subtraction method is correct. Let's assume "all three sisters are not seated together" means that no two sisters are adjacent. Using the gap method, arrange the 8 brothers in a circle in \( (8-1)! = 7! \) ways. This creates 8 gaps between the brothers. To ensure no two sisters are together, place each of the 3 sisters in a distinct gap. This can be done in \( P(8,3) \) ways. So, the number of ways = \( 7! \times P(8,3) = 7! \times 8 \times 7 \times 6 = 7! \times 336 \). \( 7! \times 336 = 5040 \times 336 = 1693440 \). Let's calculate \( 8! \times 84 \): \( 8! \times 84 = 40320 \times 84 = 3386880 \). These are different. The phrasing "all the three sisters are not seated together" means that it is not the case that (Sister Sister Sister). It does not mean (Sister NOT Sister Sister NOT Sister). Therefore, the subtraction method is correct. Total arrangements \( = 10! \) Arrangements where the 3 sisters are together \( = 8! \times 3! \) Arrangements where the 3 sisters are NOT together \( = 10! - 8! \times 3! \) \( = 10 \times 9 \times 8! - 6 \times 8! \) \( = 8!(90 - 6) \) \( = 8! \times 84 \) This matches option (D). The ambiguity of the phrase "not together" is important here. Usually, "not together" means "at least one is separate" from the group. If it means "no two adjacent", the question would usually be phrased as "no two sisters sit together".