# Permutations and Combinations

#### chrisyuen

1. The problem statement, all variables and given/known data

7 distinct flags are hoisted in a post. Find the number of ways of arranging them if
2 of the flags must be separated. (The answer is 3600)

2. Relevant equations

Permutation: n!
Circular Permutation: (n-1)!

3. The attempt at a solution

1: Flag 1, 2: Flag 2:

The arrangements should be:

1 X 2 X X X X
X 1 X 2 X X X ...

2 X 1 X X X X
X 2 X 1 X X X ...

1 X X 2 X X X
X 1 X X 2 X X ...

2 X X 1 X X X
X 2 X X 1 X X ...

......

I don't know how to calculate the solution provided.

Can anyone tell me how to solve the above problem?

Thank you very much!

#### CompuChip

Homework Helper
I don't know if this is the way you are supposed to find, but for me a simple counting argument always works.

Suppose that the first flag is not placed in the first or last position. How many possibilities are there? For each choice, how many possibilities are there to place the second flag? For each fixed placement of flag 1 and 2, how many possibilities are there to arrange the remaining 5?

#### chrisyuen

I don't know if this is the way you are supposed to find, but for me a simple counting argument always works.

Suppose that the first flag is not placed in the first or last position. How many possibilities are there? For each choice, how many possibilities are there to place the second flag? For each fixed placement of flag 1 and 2, how many possibilities are there to arrange the remaining 5?
1) (6 x 1 x 5 x 4 x 3 x 2 x 1) x 5 ?

2) Don't know

3) 5! ?

#### CompuChip

Homework Helper
3) is right. For any placement of the two marked flags, you have 5! possibilities to arrange the rest. So this will give you 5! x (number of placements for the first two flags).

For 1) you are thinking much too complicated. I have 7 empty places, of which there are five which have two neighbours (I am excluding the case where flag 1 is in the first or last position for a moment). I want to put the first flag in one of them. How many ways are there in which I can do that.
Then how many spots out of the 7 are still allowed for the second flag?

#### davieddy

How many arrangements have flag 1 adjacent to flag 2?

#### chrisyuen

3) is right. For any placement of the two marked flags, you have 5! Possibilities to arrange the rest. So this will give you 5! X (number of placements for the first two flags).

For 1) you are thinking much too complicated. I have 7 empty places, of which there are five which have two neighbours (i am excluding the case where flag 1 is in the first or last position for a moment). I want to put the first flag in one of them. How many ways are there in which i can do that.
Then how many spots out of the 7 are still allowed for the second flag?
1) 5 ?

#### chrisyuen

How many arrangements have flag 1 adjacent to flag 2?
Treating Flag 1 and Flag 2 as one unit, there are (5+1)! arrangements available.

However, inside that unit, there are 2! arrangements available.

Totally, we have (5+1)! x 2! arrangements?

#### chrisyuen

Treating Flag 1 and Flag 2 as one unit, there are (5+1)! arrangements available.

However, inside that unit, there are 2! arrangements available.

Totally, we have (5+1)! x 2! arrangements?
davieddy, is that the solution should be 7! - (5+1)! x 2! = 3600?

Thank you very much!

#### davieddy

Well done.
Compuchip was trying to guide you through the direct
approach whiich you were trying originally. It gives:

(5*4 +2*5) * 5! = 3600

Not really too complicated, but messier.

David

#### davieddy

Number of ways of locating flags 1 and 2 non adjacent:

5*4 + 2*5 = 7*6 - 6*2 = 30

Nothing in it really!
Perhaps there are further different ways of counting it.

David

BTW * is used in computing/math instead of x for a reason
too obvious to mention.

Last edited:

#### chrisyuen

Number of ways of locating flags 1 and 2 non adjacent:

5*4 + 2*5 = 7*6 - 6*2 = 30

Nothing in it really!
Perhaps there are further different ways of counting it.

David

BTW * is used in computing/math instead of x for a reason
too obvious to mention.
Actually, I don't understand why you can get 5*4 + 2*5 OR 7*6 - 6*2.

And why does the first flag not place in the first or last position?

#### davieddy

Actually, I don't understand why you can get 5*4 + 2*5 OR 7*6 - 6*2.

And why does the first flag not place in the first or last position?
(7*6 - 6*2)*5! is what you worked out correctly earlier.

If flag one is not at one end, then there are 4 places to put flag 2
He (Compuchip) went on to ask how many ways can you place flag 2
with flag 1 at an end. Answer 2"5.

#### chrisyuen

I got it!

Thank CompuChip & davieddy very much!

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