# Permutation

• Jan 11th 2013, 05:23 AM
dedust
Permutation
hello...I need help with this probelm :

How many ways are there for eight men and five women to stand in a line so that no two women stand next to each other? [Hint: First position the men and then consider possible positions for the women.]

This is a problem from rossen's discrete math... the answer is 609,638,400, but i can't get the same answer...
• Jan 11th 2013, 05:36 AM
Plato
Re: Permutation
Quote:

Originally Posted by dedust
How many ways are there for eight men and five women to stand in a line so that no two women stand next to each other? .

$.\underline{~~}|\underline{~~} \underline{~~}|\underline{~~}\underline{~~}| \underline{~~}\underline{~~}| \underline{~~}\underline{~~}|\underline{~~} \underline{~~}|\underline{~~}\underline{~~}| \underline{~~}\underline{~~}|\underline{~~}.$

The eight men create nine places to place the five women.

$\binom{9}{5}(8!)(5!)$
• Jan 11th 2013, 05:54 AM
dedust
Re: Permutation
" The eight men create nine places " thank you plato (Bow)