Statistics Permutations and Combinations

• resurgance2001
In summary, the number of ways for 8 people to sit in 12 seats in a cinema if there are no restrictions is 19,958,400. If Mary and Francis cannot sit in seats next to each other, the number of ways is 3,326,400. If all 8 people must sit together with no empty seats, the number of ways is 321,600. There are multiple methods to solve this problem, including counting directly and using combinations and permutations formulas.
resurgance2001

Homework Statement

The back row of a cinema has 12 seats, all of which are empty. A group of 8 people including Mary and Francis, sit in this row.

Find the number of ways they can sit in these 12 seats if

a) There are no restrictions
b) Mary and France's do not sit in seats which are next to each other
C) All 8 people sit together with no empty seats.

Homework Equations

nCr = n!/r! (n-r)! nPr = n! (n-r)!

The Attempt at a Solution

a) There are no restrictions
12P8 = 19,958,400

b) This is where I get stuck. I tried to calculate the number of arrangements where Mary and Frances are sitting next to each and then subtract this from the answer to a). [/B]

I reason that there are 11 place where the two women can sit next to each and that for each place there are two ways they can sit.

So 11 X 2 = 22

Then the remaining six people could sit in any of the 10 remaining seats 10P6 ways

This gives a totals of 22 X 5 X 6 X 7 X 8 X 9 X 10 = 3,326, 400 which is more than I found in part a). So this has to be incorrect.

c) There are 5 X 8! ways = 321,600
Which I think might be correct.

resurgance2001 said:

Homework Statement

The back row of a cinema has 12 seats, all of which are empty. A group of 8 people including Mary and Francis, sit in this row.

Find the number of ways they can sit in these 12 seats if

a) There are no restrictions
b) Mary and France's do not sit in seats which are next to each other
C) All 8 people sit together with no empty seats.

Homework Equations

nCr = n!/r! (n-r)! nPr = n! (n-r)!

The Attempt at a Solution

a) There are no restrictions
12P8 = 19,958,400

b) This is where I get stuck. I tried to calculate the number of arrangements where Mary and Frances are sitting next to each and then subtract this from the answer to a). [/B]

I reason that there are 11 place where the two women can sit next to each and that for each place there are two ways they can sit.

So 11 X 2 = 22

Then the remaining six people could sit in any of the 10 remaining seats 10P6 ways

This gives a totals of 22 X 5 X 6 X 7 X 8 X 9 X 10 = 3,326, 400 which is more than I found in part a). So this has to be incorrect.

c) There are 5 X 8! ways = 321,600
Which I think might be correct.

Please do not use bold fonts: it looks like you are yelling at us.

Amyway, one way to deal with (b) would be a not-quite-brute-force method: let ##i## be the seat number of Mary and ##j## the seat number of Francis. For any acceptable pair ##(i,j)## the remaining 10 unoccupied seats can be filled in any way by the remaining 6 people, and there are ##{}_{10}P_6 =151,200## different ways of doing that.

If ##i = 1## we must have ##j \in \{ 3, \ldots, 12 \}##, so there are ##12-3+1=10## acceptable ##j##-locations. If ##i = 2## we must have ##j \in \{ 4, \ldots, 12\},## so there are ##12-4+1 = 9## acceptable ##j##-positions. If ##i=3## we must have ##j \in \{1,5, \ldots, 12\}, ## so there are ##1+12-5+1 = 9## acceptable ##j##-positions. Continue like that for ##i = 4,5, \ldots, 12,## then sum everything

Last edited:
Alternatively, you could count the number of ways that the two could sit together.

Thanks - sorry about the bold font. It just picked it up from the template - I should have gone back to change it.

I actually found that the way I did it got a correct answer. I had made a mistake in my calculation for part a)

I will try your method also.

The make scheme just said 11P7 x 2 for the number of ways with the two women sat next to each other.

That's obviously a very succinct and quick way of doing it, but I am struggling a bit to see how they arrived at that answer unless they treated the two women (and the two seats they are sitting in) as a single unit - the multiplying it by two because there are two ways that the women could sit in each arrangement.

Thanks

PeroK said:
Alternatively, you could count the number of ways that the two could sit together.

Perok - that was what I was trying to do. I think I can see now that the answer to that question is 11P7 x 2 and then we subtract that from the answer to part a) Thanks

resurgance2001 said:
Perok - that was what I was trying to do. I think I can see now that the answer to that question is 11P7 x 2 and then we subtract that from the answer to part a) Thanks

I think you were right all along with your method, but you miscalculated somewhere.

PS it should be just as easy to count this directly (for part b). Think about where Mary sits: there at two cases: at an end seat or not. Then think about the options for Frances. Then the rest.

PeroK said:
PS it should be just as easy to count this directly (for part b). Think about where Mary sits: there at two cases: at an end seat or not. Then think about the options for Frances. Then the rest.

See post #2.

resurgance2001 said:
Thanks - sorry about the bold font. It just picked it up from the template - I should have gone back to change it.

I actually found that the way I did it got a correct answer. I had made a mistake in my calculation for part a)

I will try your method also.

The make scheme just said 11P7 x 2 for the number of ways with the two women sat next to each other.

That's obviously a very succinct and quick way of doing it, but I am struggling a bit to see how they arrived at that answer unless they treated the two women (and the two seats they are sitting in) as a single unit - the multiplying it by two because there are two ways that the women could sit in each arrangement.

Thanks

Yes, that is exactly what they have done to get ##2 \times {}_{11}P_7##.

1. What is the difference between permutations and combinations in statistics?

Permutations and combinations are both methods of arranging objects in a specific order. The main difference is that permutations involve arranging all objects in a specific order, while combinations only involve selecting a subset of objects without regard to their order. In other words, permutations are concerned with arranging objects in a specific sequence, while combinations are concerned with selecting objects without regard to their sequence.

2. How do you calculate the number of permutations of a set of objects?

The formula for calculating permutations is n!/(n-r)!, where n is the total number of objects and r is the number of objects being selected. This is because permutations involve arranging all objects in a specific order, so the number of possible arrangements is equal to the total number of objects factorial divided by the number of objects being selected factorial.

3. Can permutations and combinations be used in real-life situations?

Yes, permutations and combinations are often used in real-life situations, such as when calculating the number of possible outcomes in a game or the number of ways to arrange seating at a dinner party. They are also commonly used in statistics to calculate probabilities and make predictions.

4. How do you calculate the number of combinations of a set of objects?

The formula for calculating combinations is n!/r!(n-r)!, where n is the total number of objects and r is the number of objects being selected. This is because combinations only involve selecting a subset of objects without regard to their order, so the number of possible combinations is equal to the total number of objects factorial divided by the number of objects being selected factorial and the number of objects not being selected factorial.

5. Are there any real-life examples where permutations and combinations are used together?

Yes, there are many real-life examples where permutations and combinations are used together. One common example is in lottery games, where the order of the numbers drawn does not matter (combinations), but the total number of possible combinations is calculated (permutations). Another example is in creating unique passwords, where the order of the characters does not matter (combinations), but the number of possible combinations is calculated (permutations).

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