SOLUTION: In how many ways 5 girls & 5 boys can be arranged if at least 2 girls sit together? I know one method 10! - 5!x6! but i want to it by another method so please help me for

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First we figure the number of ways that anybody can sit anywhere:

That's 10!

Now we figure the number of ways in which no two girls can sit 
together. These are:

1. this way, alternating seats with a boy on the far left and
a girl on the far right:

B G B G B G B G B G

G B G B G B G B G B

Each of those is (5!)²

That's 2×(5!)²

2. these ways, with one pair of boys together

G (B B) G B G B G B G

G B G (B B) G B G B G

G B G B G (B B) G B G

G B G B G B G (B B) G

Each of those 4 ways are also (5!)²

So that's 4×(5!)²

So there are a total of 

2×(5!)² + 4×(5!)²  ways in which no girls sit 
together. So we subtract 6×(5!)² from the total number of 
ways there are to seat anybody anywhere. The end result will
count only the ways in which two or more girls sit together.

10! - 6(5!)² = 3628800 - 6(120)² = 3628800 - 6(14400) =

3628800 - 86400 = 3542400 

Edwin