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Wouldn't it just be the number of spaces around the women choose 6? |-|-|-|-|-|-|-|-|-|-| if this picture makes sense... should be 11, right? one between each pair of women then one on the far left and one on the far right--[[User:Spfeifer|Spfeifer]] 20:19, 4 February 2009 (UTC) | Wouldn't it just be the number of spaces around the women choose 6? |-|-|-|-|-|-|-|-|-|-| if this picture makes sense... should be 11, right? one between each pair of women then one on the far left and one on the far right--[[User:Spfeifer|Spfeifer]] 20:19, 4 February 2009 (UTC) | ||
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| + | Spfeifer has the right approach but don't forget to multiply that by 10! for the combinations of women and by 6! for the combinations of men--[[User:Kfox|-Kristen]] 20:30, 4 February 2009 (UTC) | ||
Revision as of 15:30, 4 February 2009
Any hints for this problem?
I used the ---|---| method. I placed the 10 women (-) and then the men (|) and then used 16 choose 6. Im not sure if this is absolutely correct but it was the direction I went.
I started to do the same thing, but is the requirement met that no more than two women stand next to each other? I guess I'm still a little confused. --Rhollowe 16:44, 4 February 2009 (UTC)
The requirement is that no two men stand next to each other, so I believe it would as long as all the (|)'s are used.
Wouldn't it just be the number of spaces around the women choose 6? |-|-|-|-|-|-|-|-|-|-| if this picture makes sense... should be 11, right? one between each pair of women then one on the far left and one on the far right--Spfeifer 20:19, 4 February 2009 (UTC)
Spfeifer has the right approach but don't forget to multiply that by 10! for the combinations of women and by 6! for the combinations of men---Kristen 20:30, 4 February 2009 (UTC)
