Statistics Permutations and Combinations

[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.

 

[Ray Vickson]

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

 

[PeroK]

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

 

[resurgance2001]

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

 

[resurgance2001]

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

 

[PeroK]

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.

 

[PeroK]

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.

 

[Ray Vickson]

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.

 

[Ray Vickson]

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 $2 \times {}_{11}P_7$.