5 men and 5 women are to be seated in a row of 10 seats. How many ways can the men not sit together and the women not sit together?
[Math] How many ways can the men not sit together and the women not sit together
combinatoricsnumber theory
Related Solutions
The men can go first, or the women can go first. Two options means multiply by two.
For another example, what if there is one man and one woman? How many ways can they line up? Well $1!\times 1!\times 2!$
Two circular seating arrangements are considered to be distinguishable if the relative order of the people differs.
In how many ways can six men sit at a round table?
Method 1: Consider the diagram below.
We can describe a particular seating arrangement by describing the order of the men as we proceed counterclockwise (anticlockwise) around the circle. Notice, however, that we can describe this particular seating arrangement in six ways, depending on whether we start with A, B, C, D, E, or F. Thus, this particular circular arrangement corresponds the six linear arrangements ABCDEF, BCDEFA, CDEFAB, DEFABC, EFABCD, FEABCD. More generally, each circular arrangement of the six men corresponds to six linear arrangements, corresponding to the six possible starting points as we proceed counterclockwise around the circle. Since six men can be arranged in a row in $6!$ ways, there are $$\frac{6!}{6} = 5!$$ distinguishable circular arrangements of the men.
More generally, we can arrange $n \geq 1$ distinct objects in a circle in $$\frac{n!}{n} = (n - 1)!$$ ways since each circular arrangement corresponds to $n$ linear arrangements, one for which each of the $n$ starting points as we proceed counterclockwise around the circle.
Method 2: We consider arrangements relative to the first man we sit at the table.
We seat man A. We use him as our reference point. As we proceed counterclockwise around the circle, we can seat the remaining five men in $5!$ orders relative to man A.
More generally, given a set of $n$ distinct objects, we can arrange them in $(n - 1)!$ orders as we proceed counterclockwise around the circle relative to the first object we place on the circle.
In how many ways can six men and five women dine at a round table if no two women sit together.
We can seat the men in $5!$ distinguishable ways. This create six spaces, one to the immediate right of each man. To ensure that no two of the women sit in adjacent seats, we must choose five of those six spaces in which to insert a woman. Once those spaces have been selected, we can arrange the five women in those spaces in $5!$ orders. Hence, the number of possible seating arrangements is $$5! \binom{6}{5} \cdot 5! = 5!P(6, 5)$$
Best Answer
If you want to find the number of ways to arrange men and women so that we have no men or women sitting consecutively we get
$$2\cdot5!\cdot5!$$ Since There $5!$ ways to arrange the males skipping a seat in between and $5!$ ways to arrange the females in the remaining $5$ spots. Then there are $2$ ways to start the row, either male or female.
Also if you're trying to find the number of ways to only group men together and women together (MMMMMWWWWW and WWWWWMMMMM) the number is the same, $2\cdot5!\cdot5!$