Number of ways to arrange 9 people in 5 spots, given conditions

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In summary, the total number of ways to arrange 9 people in 5 spots is 9P5, or 15,120. The conditions for arranging them are that each spot must have one person and all 9 people must be included. This problem differs from a traditional permutation problem because not all people can be included. The order of people in each spot cannot be changed. To solve this problem efficiently, use the formula nPr = n! / (n-r)! with n = 9 and r = 5.
  • #1
leo255
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Homework Statement



Nine people (Ann, Ben, Cal, Dot, Ed, Fran, Gail, Hal, and Ida) are in a room. Five of them stand in a row for a picture. In how many ways can this be done if

(e) Hal or Ida (but not both) are in the picture?

(f) Ed and Gail are in the picture, standing next to each other?
  1. (g) Ann and Ben are in the picture, but not standing next to each other?

Homework Equations

The Attempt at a Solution


[/B]
Given no conditions, the answer would obviously be P(9, 5).

For part "e," one of them can be in the picture. Take both out of the entire pool, and one out of the group, taking pictures. This gives me P(7, 4).

For part "f," I can take two out of both groups (i.e. total and group picture), that leaves me P(7, 3).

For part "g," I don't know how to calculate how they will be ordered, other than to give an incomplete answer of P(7, 3).

I know that all of my answers are wrong.
 
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  • #2
leo255 said:

Homework Statement



Nine people (Ann, Ben, Cal, Dot, Ed, Fran, Gail, Hal, and Ida) are in a room. Five of them stand in a row for a picture. In how many ways can this be done if

(e) Hal or Ida (but not both) are in the picture?

(f) Ed and Gail are in the picture, standing next to each other?
  1. (g) Ann and Ben are in the picture, but not standing next to each other?

Homework Equations

The Attempt at a Solution


[/B]
Given no conditions, the answer would obviously be P(9, 5).

For part "e," one of them can be in the picture. Take both out of the entire pool, and one out of the group, taking pictures. This gives me P(7, 4).

For part "f," I can take two out of both groups (i.e. total and group picture), that leaves me P(7, 3).

For part "g," I don't know how to calculate how they will be ordered, other than to give an incomplete answer of P(7, 3).

I know that all of my answers are wrong.

How do you know you answers are all wrong; what do you think are the correct answers?

Anyway, what do you want us to do? Confirm your suspicions? Show you how to do the questions (which would violate PF rules)?
 
  • #3
If the notation P(a,b) is the number of combinations of b elements among a, then your answers are incorrect but not stupid.
You forgot to account for a number of configurations in each case. So the correct answers are a multiple of your answers.

EDIT: your answers to e,f,g
 
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  • #4
Ray Vickson said:
How do you know you answers are all wrong; what do you think are the correct answers?

Anyway, what do you want us to do? Confirm your suspicions? Show you how to do the questions (which would violate PF rules)?

I actually have answers, but they do me no good if I can't figure this on my own. I honestly try not to look at the answers, but here they are:

e is 2 * 5 * P(7, 4)
f is 2 * 4 * P(7, 3)
g is 5 * 4 * P(7,3) - 2 * 4 * P(7, 3)

I obviously got one part of them correct. To be honest, I probably don't even need help with g, since it just takes from f (I understand what 5 * 4 * P(7,3) means), but I am really confused about e and f, and how these numbers are gotten.

Specifically for e, if only one person is in the picture, I could see 5 * P(7, 4), but I don't know why we would need a 2. I am clueless about where the 2 * 4 comes from for f.
 
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  • #5
Ok, so answering my own question, I now see where the 2 comes from for e. We need to account for the possibility of one out of the two being in the picture. Still working on f.
 
  • #6
geoffrey159 said:
If the notation P(a,b) is the number of combinations of b elements among a, then your answers are incorrect but not stupid.
You forgot to account for a number of configurations in each case. So the correct answers are a multiple of your answers.

EDIT: your answers to e,f,g
leo255 said:
I actually have answers, but they do me no good if I can't figure this on my own. I honestly try not to look at the answers, but here they are:

e is 2 * 5 * P(7, 4)
f is 2 * 4 * P(7, 3)
g is 5 * 4 * P(7,3) - 2 * 4 * P(7, 3)

I obviously got one part of them correct. To be honest, I probably don't even need help with g, since it just takes from f (I understand what 5 * 4 * P(7,3) means), but I am really confused about e and f, and how these numbers are gotten.

Specifically for e, if only one person is in the picture, I could see 5 * P(7, 4), but I don't know why we would need a 2. I am clueless about where the 2 * 4 comes from for f.

For the first one, the total number of arrangements = arrangements that have Ida but not Hal + arrangements that have Hal but not Ida. These two numbers are the same, and the first one = 5*P(7,4). Note that adding the two numbers together is OK (no "double counting", for example) because the total number of pictures is the number containing Ida plus the number containing Hal. For the second one, let E and G form a new "superperson", like two persons glued together. You need to locate that one superperson plus three others from the remaining 7; the remaining three can be arranged in P(7,3) ways, then the superperson inserted in 4 ways, then be separated again into E and G in two possible orders.
The last one looks at the number of pictures having both A and B in them, then subtracts the number where A and B are next to each other.
 
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  • #7
Thanks! I think I got it. I was really getting confused by the '4,' but see that those are the only possibilities (i.e. 1,2; 2,3; 3, 4; 4, 5) - the fifth option would be 5, 6, and that would not exist in this context. The 2, as you said, is just where they happen to be positioned.
 
  • #8
I disagree with these answers.

Let's take (e) :
P(7,4) is the number of sets of 5 person in the pool containing Hal but not Ida (or Ida but not Hal). In each of these sets, people can be ordered in ##5!## different ways. So the answer should be ##2\times (5!)\times P(7,4)## and not ##2\times 5\times P(7,4)##.
 
  • #9
geoffrey159 said:
I disagree with these answers.

Let's take (e) :
P(7,4) is the number of sets of 5 person in the pool containing Hal but not Ida (or Ida but not Hal). In each of these sets, people can be ordered in ##5!## different ways. So the answer should be ##2\times (5!)\times P(7,4)## and not ##2\times 5\times P(7,4)##.

The notation ##P(n,m)## means the number of permutations of ##m## things chosen from a group of ##n##, so already includes all the different "orders". In contrast, the number of combinations of ##m## things chosen from ##n## things is ##C(n,m) = P(n,m)/m!##, because when disregarding the order the same ##m## things can be ordered in ##m!## different ways.

So, ##2 \times 5 \times P(7,4)## is correct, because for each of the ##P(7,4)## permutations of the 4 things we can insert the 5th thing into 5 different places.
 
  • #10
Oh Ok, I asked the the OP what was the meaning of this notation but did not reply about that so I assumed it was the number of combinations. So I agree then !
 
  • #11
Sorry for not making all of that clear. Again, thanks guys for your help!
 

1. How many ways can 9 people be arranged in 5 spots?

The total number of ways to arrange 9 people in 5 spots is 9P5, which is equal to 15,120.

2. What are the conditions for arranging the 9 people in 5 spots?

The conditions for arranging the 9 people in 5 spots are that each spot must have exactly one person and all 9 people must be included in the arrangement.

3. How is this problem different from a traditional permutation problem?

This problem differs from a traditional permutation problem because there are more people than spots available, so not all people can be included in the arrangement.

4. Can the order of the people in each spot be changed?

No, the order of the people in each spot cannot be changed as the conditions specify that each spot must have exactly one person.

5. How can I solve this problem efficiently?

This problem can be solved efficiently by using the formula for permutations with restrictions, which is nPr = n! / (n-r)!, where n is the total number of items and r is the number of items being arranged. In this case, n = 9 and r = 5, so the equation becomes 9! / (9-5)! = 15,120.

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