Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 6 - Section 6.3 - Permutations and Combinations - Exercises - Page 413: 13

Answer

$2\times(n!)^2$

Work Step by Step

We have 2 options as we can start the row by either men or women. Let's say we start with a woman. Now because men and women alternate, women must occupy all the odd places and men must occupy all the even places. So women can shuffle among themselves(at odd places) in $n!$ ways and so can men. So we have $n! \times n!$ possibilities. Similarly, if we start with a man, we have $n! \times n!$ possibilities. So, in total, we have $n! \times n!+n! \times n!$ = $2\times(n!)^2$ total possible arrangements.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.