In how many ways can six girls and two boys be rearranged in a row. 

(1) without ristriction 


Choose the 1st person as any of the 8 people
Choose the 2nd person as any of the 7 remaining unchosen people.
Choose the 3rd person as any of the 6 remaining unchosen people.
Choose the 4th person as any of the 5 remaining unchosen people.
Choose the 5th person as any of the 4 remaining unchosen people.
Choose the 6th person as any of the 3 remaining unchosen people.
Choose the 7th person as either of the 2 remaining unchosen people.
Choose the 8th person as the only 1 remaining unchosen person.

8*7*6*5*4*3*2*1 = 8! = 40320 ways
------------------------------------------
(2) such that the boys are not together.

From the 40320 from part (1) we must subtract the number of ways
the boys can be together.  They can be considered as a "pair of boys"
in 2 ways, with Tom on Dick's left, and with Dick on Tom's left.

With Tom on Dick's left:

As above except now we have 7 "things" to arrange,

6 girls and 1 "pair of boys".

That will be 7*6*5*4*3*2*1 = 7! = 5040 ways.

With Dick on Tom's left:

That's also 5040 ways.

So that's 2*5040 or 10080 to subtract from the 40320.

40320 - 10080 = 30240 ways. 

Edwin