SOLUTION: The following problems deal with permutations of the eight letters A, B, C, D, E, F, G and H. a. What is the number of 8-permutations if A and B have to be among the first three

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Question 890824: The following problems deal with permutations of the eight letters A, B, C, D, E, F, G and H.
a. What is the number of 8-permutations if A and B have to be among the first three letters?



b. What is the number of 6-permutations if ‘FG’ should not be at the end?




c. What is the number of 5-permutations containing A, B and C but not adjacent to one another?

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
The following problems deal with permutations of the eight letters A, B, C, D, E, F, G and H.
a. What is the number of 8-permutations if A and B have to be among the first three letters? 
Choose the position to place the A as any of the first 3 positions.
Choose the position to place the B as either of the two remaining positions
of the first three.
Choose the position to place the C as any of the remaining 6 positions.
Choose the position to place the D as any of the remaining 5 positions.
Choose the position to place the E as any of the remaining 4 positions.
Choose the position to place the F as any of the remaining 3 positions.
Choose the position to place the G as either of the remaining 2 positions.
Choose the position to place the H as the only 1 remaining position.

That's 3×2×6×5×4×3×2×1 = 6×6! = 6×720 = 4320

b. What is the number of 6-permutations if ‘FG’ should not be at the end? 
First we calculate the number of permutations which is 8! = 40320

Now we will calculate the number of unwanted permutations with FG at the end.

Choose the 7th position to place the F.
Choose the 8th position to place the G.
Choose the position to place the A as any of the first 6 positions.
Choose the position to place the B as any of the 5 remaining positions
of the first six.
Choose the position to place the C as any of the 4 remaining positions
of the first six.
Choose the position to place the D as any of the 3 remaining positions
of the first six.
Choose the position to place the E as either of the 2 remaining positions
of the first six.
Choose the position to place the H as the 1 remaining positions
of the first six.

That's 1×1×6×5×4×3×2×1 = 6! = 720

Now we subtract those 720 from the 40320, getting 40320-720 = 39600

c. What is the number of 5-permutations containing A, B and C but not adjacent to one another? 
They must go like this:

%28matrix%285%2C1%2C%0D%0AEither%2CA%2CB%2Cor%2CC%29%29%28matrix%287%2C1%2C%0D%0AEither%2CD%2CE%2CF%2CG%2Cor%2CH%29%29%28matrix%285%2C1%2C%0D%0AEither%2CA%2CB%2Cor%2CC%29%29%28matrix%287%2C1%2C%0D%0AEither%2CD%2CE%2CF%2CG%2Cor%2CH%29%29%28matrix%285%2C1%2C%0D%0AEither%2CA%2CB%2Cor%2CC%29%29

Choose a letter for the 1st position in 3 ways  (either A,B, or C)
Choose a letter for the 3rd position either of 2 ways
Choose a letter for the 5th position 1 way.
Choose a letter for the 2nd position any of 5 ways. (D,E,F,G, or H)
Choose a letter for the 4th position any of 4 ways.  

3×2×1×5×4 = 120

Edwin