How many permutations of ABCDEF have A appearing before F?

Question

How many permutations of ABCDEF have A appearing before F?

I know that combinations are used to solve this problem, yet I do not know how.

also,

How many permutations of the letters ABCDEF have A and F separated by at least one letter?

would this be (Total # of Permutations) - (# Permutations with A and F touching each other)?

Answer 1

The easy way to think about this would be to:

6 distinct letters have 6! (that's factorial 6 = 6*5*....*1) ways of being arranged. Either A is before F or F is before A. and, lol, since these are fairly equal letters, we can assume safely that half of these would have A before F and the other half F before A. Therefore the answer is 6!/2 = 360 ways.

For the second part you need to tell me whether seperation by one letter also means that the last letter is considered adjacent to the first letter (ring formation). Then the number is higher.

Answer 2

A F _ _ _ _

so, there is only one A for the first letter and one F for the second letter. Since two letters are already taken there are 4 letters left

1 x 1 x 4 x 3 x 2 x 1 = 24

BUT, A and F does not have to be in the first place and second place. A can be in the second place and F can be in the third place.

so, there are 5 ways you can rearrange AF within the remaining 4 letters

so 5 x 24 = 120 ways (if you're talking about F must follow A)




Separate at least 1 letter:
A _ F _ _ _ (4 ways to rearrange A and F)

1 x 4 x 1 x 3 x 2 x 1 = 24
24 x 4 = 96

A _ _ F _ _ (3 ways to arrange A and F)

1 x 4 x 3 x 1 x 2 x 1 = 24
24 x 3 = 72 ways

A _ _ _ F _ (2 ways to rearrange A and F)
1 x 4 x 3 x 2 x 1 x 1 = 24
24 x 2 = 48 ways

A _ _ _ _ F (only 1 ways to rearrange A and F)
1 x 4 x 3 x 2 x 1 x 1 = 24

96 + 72 + 48 + 24 = 240 ways

BUT A and F can be rearrange within themselves
240 x 2 = 480 ways

Answer 3

First question:
there are 6*5*4*3*2*1 of arranging the letters, that is, 720.

Obviously exactly half of these will have A before F, so half of 720 is 360.
.
If we look at ABCDEF then A and F are separated from 4 of 5, B, C, D, and E from 3 of 5. So for your second question, we have 4/5 * 120 * 2 + 3/5 * 120 * 4 = 192 + 288 = 480.
.

Answer 4

Let us first observe the # of permutations on 6 letters is 720 and 5 letters is 120.

For the first problem, half the total permutations have A before F so the answer is 720/2 = 360.

For the second, it is (as you suggest)
720 - # of permuations with A and F touching.

Now if A and F are touching, it is in the order AF or FA.

Suppose it is AF. Then AF can be thought of as 1 letter and not two. So the # of permuations is the same as 5 letters or 120. And there are also 120 permuations that have FA.

So the answer is 720 - 2 * 120 = 480.

Answer 5

The number of ways to place A and F such that A always comes before F (though not necessarily immediately before F) is 5+4+3+2+1 = 15. They look like this, with x indicating a character other than A or F:

AFxxxx
AxFxxx
AxxFxx
AxxxFx
AxxxxF
xAFxxx
xAxFxx
xAxxFx
xAxxxF
xxAFxx
xxAxFx
xxAxxF
xxxAFx
xxxAxF
xxxxAF

Now we find the number of ways to place the remaining four letters (where the x characters are). That is simply 4! = 4*3*2*1 = 24. The total number of ways to arrange it is the product of these two numbers: 15*24 = 360 arrangements.