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The way I did it was counted up the possible permutations for the women which is P(10,10) and then like Spfeifer said counted out 11 possibilites for the men to be and took P(11,6) 6 for the six men. so the final answer would be P(10,10)x P(11,6) --[[User:Krwade|Krwade]] 20:52, 4 February 2009 (UTC) | The way I did it was counted up the possible permutations for the women which is P(10,10) and then like Spfeifer said counted out 11 possibilites for the men to be and took P(11,6) 6 for the six men. so the final answer would be P(10,10)x P(11,6) --[[User:Krwade|Krwade]] 20:52, 4 February 2009 (UTC) | ||
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| + | -I don't want to get out my calculator to see if any of these are correct, but I can't seem to find any overcount in my solution, and I think that it is exhaustive: I first start by putting the forced positions down. Let X's be girls and O's be guys. Therefore the forced positions are: O X O X O X O X O X O X O. (none of these are labeled yet, and we still have five girls to place. We have 7 "boxes" where we can place these remaining girls: by any of the other O's. Therefore we have (7 + 5 - 1 choose 7) possibilities. Then multiply by 10! and 6! to give them all indices, giving us an answer of: | ||
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| + | <math>10! * 6! * \binom{11}{7}</math> | ||
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| + | Anyone see problems? Please let me know. -mkburges | ||
Revision as of 18:06, 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)
The way I did it was counted up the possible permutations for the women which is P(10,10) and then like Spfeifer said counted out 11 possibilites for the men to be and took P(11,6) 6 for the six men. so the final answer would be P(10,10)x P(11,6) --Krwade 20:52, 4 February 2009 (UTC)
-I don't want to get out my calculator to see if any of these are correct, but I can't seem to find any overcount in my solution, and I think that it is exhaustive: I first start by putting the forced positions down. Let X's be girls and O's be guys. Therefore the forced positions are: O X O X O X O X O X O X O. (none of these are labeled yet, and we still have five girls to place. We have 7 "boxes" where we can place these remaining girls: by any of the other O's. Therefore we have (7 + 5 - 1 choose 7) possibilities. Then multiply by 10! and 6! to give them all indices, giving us an answer of:
$ 10! * 6! * \binom{11}{7} $
Anyone see problems? Please let me know. -mkburges
