# Variations question

1. Feb 16, 2008

### Physicsissuef

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

On how many ways, 8 different people can sit on 12 seats?

2. Relevant equations

$$V\stackrel{k}{n}$$

3. The attempt at a solution

I tried to solve this task by using the formula of variations without repetition.

$$V\stackrel{8}{12}$$

But I sow in my textbook results, and it says that it should be: 8! * $$V\stackrel{8}{12}$$

I don't know what is correct actually.

2. Feb 16, 2008

### HallsofIvy

Staff Emeritus
I have absolutely no idea what $V_n^k$ means!

I suspect that that it is what I would call $_nC_k$ or $\left(\begin{array}{c}n \\ k \end{array}\right)$, the binomial coefficient. That is the number of ways we can pick a group of 8 chairs out of 12 for the people to sit on. That, however, does not include the number of orders in which those 8 people can sit on the same 8 chairs. Since there are 8! orders for 8 people, that would be $8!_12C_8$ or 12!/(12-8)!= 12!/4!

3. Feb 17, 2008

### Physicsissuef

$V_n^k$=n(n-1)(n-2)...(n-k)
So, 8!*$V_12^8$ will come to be, big number.
8! * 12*11*10*9*8*7*6*5

4. Feb 17, 2008

### HallsofIvy

Staff Emeritus
That is almost what I would call the number of Permutations:
$$_nP_k= \frac{n!}{(n-k)!}$$
("Almost" because what you have is
$$\frac{n1}{(n-k-1)!}$$)
and that is exactly what I gave before: k! nCk. If you text really has another 8! I have no idea why they are doing that. Perhaps it is a typo.

5. Feb 17, 2008

### Physicsissuef

I also hope so, that it is typo. Because if I get 2 people to sit on 3 seats (simplified edition of the task above), there will be 6 ways to sit.
$V_3^2$=3*2=6
or
2!*$C_3^2$=2*3=6