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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




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Topic review (newest first)

2013-03-24 22:21:29

All the forms there were obtained empirically. By looking at the data. There just was no other way that I could solve the problem,

As far as my understanding goes, it computes the number of ways you can give out n names to n people without any of them having their own name.

That is the number of derangements which is always an integer. Your sum there is not always an integer.

2013-03-24 22:13:58

I am trying to understand the logic behind the terms that's all.

Summation from j= 0 to n of (-1)^j/j! in the numerator is not getting into my head.

As far as my understanding goes, it computes the number of ways you can give out n names to n people without any of them having their own name.

Why do we need to multiply the summation in the numerator with P(no one getting drunk)?

2013-03-24 20:51:19

Look at the formula for no one getting drunk. The (k-1)! is the number of them.

Now if you tell me what you are trying to do I can suggest a good means of computing the numbers.

2013-03-24 20:42:28

and why did we have to time that by (k-1)!?

2013-03-24 20:41:22

That is quite okay, thank you.
Although I do not understand the 2nd summation in the numerator. What does it compute? the probability that n people (who are gonna get drunk) don't have each other's name?

2013-03-24 20:22:59

Then please hold on, it will take awhile.

Sorry, but the closed form is gigantic. Would a recurrence relation be okay?

2013-03-24 20:22:28

Oh yes please. Thank you big_smile.

2013-03-24 20:00:52


It is no bother.

J is what is called a dummy variable. It is an index of summation.

I can try to get a closed form for the whole expression.

2013-03-24 19:55:58

Hi, sorry for bothering again, but what is j in this case?

2013-03-24 19:46:49

here is something slightly less complicated inasmuch as there is no gamma function.

2013-03-24 19:35:20

I think there's a gamma distribution in there? Unfortunately I haven't learned that yet. Would be okay if you rewrite the formula without the gamma distribution? thank you very much. smile

2013-03-24 19:25:15


I used the idea of spotting the pattern. This is in the style of the new experimental mathematics or should I say the old.

We could then work on a method to prove the formula, say by induction.

Very often in combinatorics formulas and answers are found in this manner.

For your second question, that is the simplest form I know. What do you need explained?

2013-03-24 19:17:51

bobbym wrote:


If possible, can you please write this in a simplier format? as I don't quite understand these notations. Thank you very much. big_smile

2013-03-24 19:14:54

Hi, would you be able to show me the working for b, for the general formula of expected number of drunk please? thank you very much smile.

2013-03-19 23:25:14

The programming allows me to play spot the pattern much more than without it. Basically without it the general form produced is very difficult if not impossible for all but the best of the best.

But to answer a) when stuck it is not a disgrace to list them all and do it by hand. Anything beats no answer!

Calling each person 1 to 5, the number of different ways to arrange them with no fixed points is called derangements. For 5 of anything there are 44 of them.

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