Question 1202902: If the letter E can never be first or last, how many distinct arrangements of the letters A, B, C, D, E, and
F are possible?
(A) 120
(B) 240
(C) 480
(D) 600
(E) 720
Found 3 solutions by Alan3354, greenestamps, ikleyn:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! If the letter E can never be first or last, how many distinct arrangements of the letters A, B, C, D, E, and F are possible?
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You must specify if letters can be repeated, eg BBBBBB, and do all arrangements have 6 letters, eg, is ACD acceptable?
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
"... arrangements of the letters A, B, C, D, E, and F..." means the 6 letters are used once each....
Place the letters in positions using the restrictions first.
E can't be first, so there are 5 choices for the first letter.
E can't be last, so there are 4 choices left for the last letter.
(NOTE: the solution will be the same if you choose the last letter first and then the first letter.)
After choosing the first and last letters, there are no other restrictions on the remaining 4 letters, so they can be arranged in any of 4! = 24 ways.
ANSWER: (5)(4)(24) = 480 option C
Answer by ikleyn(52818)  (Show Source):
You can put this solution on YOUR website! .
There is another way to solve the problem.
In all, we have 6 letters A, B, C, D, E, and F, and
6! = 6*5*4*3*2*1 = 720 of their possible permutations.
From it, subtract 5! = 120 permutations, where letter C is first,
and another 5! = 120 permutations, where letter C is the last.
You get the answer 720 - 120 - 120 = 480 permutations.
ANSWER. 7! - 5! - 5! = 720 - 2*120 = 480 permutations.
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