The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit t

Q: Wait, why is it $(6-1)!$ for the men? Shouldn't it be $6!$?

Great question! In a linear row, $6$ people can be arranged in $6!$ ways. But at a round table, rotating everyone by one seat doesn't create a new arrangement—it's just the same relative order. To account for this, we fix one person to 'anchor' the circle, which effectively leaves $6-1=5$ people to be arranged, hence $(6-1)!$.

Q: Why are there 6 gaps? If I have 6 men, shouldn't there be 7 gaps like in a line?

That is the classic trap! In a straight line, $n$ people create $n+1$ gaps (including the ends). But in a circle, the 'ends' meet, closing the loop. If you draw it out, you will see that 6 men in a circle create exactly 6 spaces between them. Always remember: in a circle, the number of gaps equals the number of people.

Q: Can we arrange the women first instead?

You absolutely can, but you must be careful! If you arrange the 5 women first, you would have to account for the circular symmetry of the women, and then place the men in the gaps. It is mathematically equivalent, but the Gap Method is almost always more intuitive when the constraint is 'no two X together'.

Question

The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by

Options:

(A) 7! 	imes 5!

(B) 6! 	imes 5!

(C) 30!

(D) 5! 	imes 4!

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To solve the problem of seating $6$ men and $5$ women at a round table such that no two women sit next to each other, we employ the **Gap Method**. This technique is the standard approach for handling "no two together" constraints in combinatorics.

### The Men as Anchors

First, we ignore the women and focus on the unrestricted group: the $6$ men. We seat the $6$ men around the circular table.

Because circular arrangements lack a fixed starting point, we fix one man to break the rotational symmetry. This leaves us with $5$ men to arrange in the remaining seats.

The number of ways to arrange the men is given by:
$$(6-1)! = 5! = 120$$

### The Geometry of Gaps

With the $6$ men seated, we examine the spaces between them. In a circular arrangement, the number of gaps created is exactly equal to the number of people seated.

Since there are $6$ men, there are exactly $6$ gaps available. By placing the women into these gaps, we ensure that no two women are ever adjacent, as each gap acts as a buffer.

### Placing the Women

We have $5$ women to place into $6$ available gaps. We must choose $5$ gaps out of the $6$ and arrange the women within them.

This is a permutation problem of $5$ distinct items into $6$ distinct positions, denoted as $^6P_5$. The calculation is as follows:

$$^6P_5 = \frac{6!}{(6-5)!} = \frac{6!}{1!} = 720$$

### Final Calculation

According to the **Fundamental Principle of Counting**, we multiply the number of ways to arrange the men by the number of ways to arrange the women.

The total number of ways is:
$$5! 	imes 6! = 120 	imes 720$$

Calculating the final product:
$$120 	imes 720 = 86,400$$

The total number of ways to seat the group under the given constraints is **86,400**.

Animated Solution for Mathematics - Permutations and Combinations: The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by

Select Answer:

Visualized Solution

Visualizing the Round Table

The Gap Method Strategy

Circular Permutation Formula

Arranging the 6 Men

Identifying the Gaps

Arranging 5 Women in 6 Gaps

Calculating $^{6} P_{5}$

Applying the Multiplication Principle

Final Answer & Key Takeaway

The Sigma Insight: Circular Permutations

Analyzing the Setup

The Men as Anchors

The Geometry of Gaps

Placing the Women

Final Calculation

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