Probability Problem regarding derangement. Desperately asking for help!

A jeweler received 8 watches in their respective boxes. He took them out and put them back in. What is the probability that

a-One watch in the right box

b-At most one watch in the right box

c-at least one watch in the right box.....

I knw this involves the derrangments formula, but i dont know how to use it, please teach me, and please dont stretch it, teach me and give me answers too, tomorrow is my exam:S i cant get thru this practice question

Re: Probability Problem regarding derangement. Desperately asking for help!

Quote:

Originally Posted by

**khamaar** A jeweler received 8 watches in their respective boxes. He took them out and put them back in. What is the probability that

a-One watch in the right box

b-At most one watch in the right box

c-at least one watch in the right box.....

I knw this involves the derrangments formula, but i dont know how to use it, please teach me,

I will be glad to help. We are not a tutorial service.

Therefore, you must answer the questions I ask.

Let $\displaystyle \mathcal{D}(n)$ denote the number of derangements of $\displaystyle n$ objects.

a) $\displaystyle 8\cdot\mathcal{D}(7)$. Now you explain where each of those comes from.

The we can do the next one.

Re: Probability Problem regarding derangement. Desperately asking for help!

I think that D(7) shows the number of derrangments, and then since one watch has been placed in the right box, so it could be the first, second, third......eighth ..and so 8 is multiplied....? is that how it is? 8.D(7) would show all such possible ways of placing the watches, so like to calculate the probablity we have to divide it by 8!?

and i also don't know how to calculate derangement, is there a function for it in calculators, or is there some kind of a series....??

Re: Probability Problem regarding derangement. Desperately asking for help!

Quote:

Originally Posted by

**khamaar** and i also don't know how to calculate derangement, is there a function for it in calculators, or is there some kind of a series....??

$\displaystyle \mathcal{D}(n) = n! \cdot \sum\limits_{k = 0}^n {\frac{{( - 1)^k }}{{k!}}} $

Re: Probability Problem regarding derangement. Desperately asking for help!

Iv been trying in vain to make some kind of formula for these problems since u wudnt tell me... And herez wat i got plz tell me if its right at least....If we have n objects and we want to arrange them in such a way that only m objects retain their original positions then the number of possible arrangements is.. (mCn)x[D(n-m)] ... C stands for combinations.... So plz tell me.. Is this right\?

Re: Probability Problem regarding derangement. Desperately asking for help!

Quote:

Originally Posted by

**khamaar** Iv been trying in vain to make some kind of formula for these problems since u wudnt tell me... And herez wat i got plz tell me if its right at least....If we have n objects and we want to arrange them in such a way that only m objects retain their original positions then the number of possible arrangements is.. (mCn)x[D(n-m)] ... C stands for combinations.... So plz tell me.. Is this right\?

I truly do not understand the above statement.

In reply #4, I gave you the exact formula for calculating the number of derangements. It is derived by use of inclusion/exclusion.

If we want to know the number of ways that exactly four of the eight watches end up in the correct box, the answer is:

$\displaystyle \binom{8}{4}\cdot\mathcal{D}(4).$