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Given the letters A,B,C,D,E, and F, determine the number of different ways the six letters can be arranged when the letter A is neither at the beginning nor at the end.
Leftmost position: 5 ways
Next, rightmost position: 4 ways
Next, 2nd position from the left: 4 ways
Next, 3rd position from the left: 3 ways
Next, 4th position from the left: 2 ways
Next, 5th position from the left: 1 way
How do you get that? What letter do the position refer to?
Where are you getting those numbers from? Doesn't make sense to me.
I would look at it like this. There are 6 positions for the letters, 1-6. A cannot be in position 1 or 6. So fix A in a certain position and then calculate all the arrangements when A is at that spot. Then move A to the right one and repeat. Go until A reaches position 5. Add up your answers.
Would this help?
Look at XP16a
http://img27.imageshack.us/img27/5696/015b016a.jpg
Ok. I see what was going on. I prefer to do this like the first solution. The idea is that we can find the number of ways to arrange the letters with A not in position 1 or 6 by taking the total number of permutations and subtracting the number of permutations when A is in position 1 or 6.
If you fix A to be in the 1st position, you have 5 letters left to arrange in 5 spots. The number of ways to do this is 5! (permutation formula). Same with if you fix A to be in the 6th position.
Then they found the total number of ways to arrange the six letter, 6!. Take that and subtract the permutations when A is in position 1 or 6 and you have your final answer.
I understand the first solution, but not the second, and that's why I asked for help.
Do you have any idea how the second solution is derived?