How many ways can 6 boys and 4 girls stand in a row if the girls must not be together?

Arrange all 6 boys in a row . --> ways to do this is 6!

This leaves 7 gap in between the boys to place 4 girls in. -> $_7C_4$

The girls in between the boys can have 4! ways to arrange them.

therefore, answer is $6! \cdot _7C_4 \cdot 4! $

why am i wrong ?

the right answer is $6! \cdot _7C_4 \cdot 4! + 6! \cdot _7C_2 \cdot 4! + 6! \cdot _7C_2 \cdot 4!$

Say we have girls $A,B,C,D$. The numbers that $X,Y$ are together is $2\cdot 9!$, then $X,Y$ and $Y,Z$ are together is $2\cdot 8!$. Now we use PIE.

Then at least two are together is $${4\choose 2}\cdot 2\cdot 9!-24\cdot 8! = 84\cdot 8!$$

So the answer is $$10!-84\cdot 8! = 8! (90-84) = 6\cdot 8!$$

November 24, 2018 18:38 PM

Your formula accounts for the space in between the boys, but it didn't take the spaces at the ends of the line into consideration.

September 11, 2019 16:37 PM

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