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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
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(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
