I know the correct workings. But I need to know why the ones I did won't give me the correct answer. Like, what really is wrong with them. I am following logic. Where do I fail??
In how many ways can a committee of 4 men and 4 women be seated in a row if no one is seated next to a person of the same gender?
I do it like this:
there are 4 men and 5 possible positions for the women to be seated therefore. the first woman can choose between any of the 5, the other three cannot occupy the positions at the ends of the row since that would cause two of the men to be sitting together. Hence there are only 3 possible seats for each of the three women, then 2 for the remaining two women and then 1 for the last. I multiply 4! by 5*3*2*1 therefore. The answer is not this however.
What did I do incorrectly?
It's a method in a Cambridge endorsed book that I use. But obviously I am doing something wrong, according to the author and yahooanswers.com. What do I do??????????
Assume the leftmost person is male.
Choose 1 of 4 men. Choose 1 of 4 women. Choose 1 of 3 men. Choose 1 of 3 women. Choose 1 of 2 men. Choose 1 of 2 men. Choose 1 of 1 man. Choose 1 of 1 woman.
This is 4 x 4 x 3 x 3 x 2 x 2 permutations; 576 in total.
However, the leftmost person could be a woman, in which case the same logic applies. So, there are 576 + 576 ways; 1152 in total.
sol: 1. if starting with a man, there are 4 ways to be filled for 1st seat. and also 4 ways to be filled for 2nd seat(a woman has to be seated). now to be filled for 3rd seat(by a man) ,there are now 3 ways.and continuing like this, we get the number of ways = 4*4*3*3*2*2*1*1= 576 ways
2. if we start with a woman, then get the same number of ways = 576 ways
hence the total number of possible ways = 576+576= 1152 ways