Question 1150816: The problem is: Find the number of ways in which four girls and three boys can
arrange themselves in a row so that none of the boys are together?
There are two possible solutions but only one is considered.
The first one, in which I made, is that, we can lay the three boys firstly in
the row, making it _B1_B2_B3_. Since there are 4 spaces, in each space we can
permute the girls. Thus the solution is 4!*3!=144 ways
Now the second one, we can lay the four girls before the boys, leaving blank
spaces for the boys at this configuration: _G_G_G_G_. Since there are 5 blanks,
boys can choose their positions giving it a 5P3. By multiplication, the answer
is 5P3*4!=1440 ways
And to conclude, the right answer,at most it would be considered,is the second
one, 1440 ways. My question is why the first one is not the right answer? At
what fair reason should it be accounted? Did I undercount, and/or make a gap in
my logic? Hope to shed a clear explanation on my question. Thank you.
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website!
Your first answer is incorrect because 144 is the number of ways none of the boys
AND NONE OF THE GIRLS are together. But it was not
required that none of the girls could be together.
Edwin
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