Answer to Question #338439 in Statistics and Probability for Laiba

Question

A dinner party is attended by five men and five women.

(a) How many unique ways can the 10 people sit around the table. (1)

(b) How many unique ways can the people sit around the table with men and women

alternating seats.

Answer 1

a). Consider two cases:

1). since the table is round, arrangements $1,2,3,...,10$ and $9,10,1,2,3,..., 8$ are the same. Numbers correspond to persons. We receive $9\cdot8\cdot7...1=9!$ different ways, using multiplication principle of combinatorics. 2). all seats around the table are different. Arrangements $1,2,3,...,10$ and $9,10,1,2,3,..., 8$ are different. We receive $10\cdot9\cdot8...1=10!$ different ways.

b). Consider two cases: 1). since the table is round, certain arrangements are the same due to rotations. Suppose that seats have the order: $m,w,m,w,..$, where $m$ denotes a seat for a man and $w$ denotes a seat for a woman. We receive: $5!5!$ ways using multiplication principle of combinatorics. We divide the latter by $5$ taking into account rotations and receive $4!5!$ different ways. We consider another order: $w,m,w,m,...$. We receive again $5!5!$ ways. Divide the latter by $5$ to exclude the same arrangements and receive $4!5!$ different ways. Thus, the answer is:$2\cdot4!5!$. 2). all seats around the table are different. Suppose that seats have the order: $m,w,m,w,..$, We receive: $5!5!$ different ways. We consider another order: $w,m,w,m,...$. We receive again $5!5!$ different ways. Thus, the answer is:$2\cdot5!5!$.

Answer a). 1).$9!$ different ways taking into account rotations 2). $10!$ different ways b). 1). $2\cdot4!5!$ taking into account rotations 2). $2\cdot5!5!$ different ways.