Math Is Fun Forum

This problem doesn't make sense.

Think of the "randomly distributed" deck of cards be a random permutation scramble of n objects, where n is the number of people you have in your group.  If no one gets a card with his or her own name on it, then this is a derangement.

The number of possible ways in which a deck of cards can be distributed to the party members at random is n!.  The number of derangement distributions/scrambles is subfactorial of n or !n.

Now this is where the problem doesn't make sense to me.

Okay, so this is telling me that only when there is a derangement will any names will be called out.

The phrasing of this is odd because by the quoted text above this one, no one's name will be called when there isn't a derangement.  So when the first round occurs where at least someone has his or her own card, then no body's name will be called and therefore everyone will have to drink before midnight (is this question a joke or an encrypted message saying that everyone is going to a drink vodka before midnight at a party?).

Since we weren't given a specific number of rounds the game has, and since the name card distributions are done at random, we can definitely assume that when enough rounds are executed, everyone will have to drink before midnight.

So no matter what, every round that "counts" towards determining the fate of any given players blood alcohol levels is one in which everyone will have to drink vodka.

So all I can say is the probability that no one person gets back his or her card in a single distribution round is subfactorial(n)/n!, which is a general formula which holds true for any n.

So if we were given that there were 15 rounds in the game, and we had say, 5 players, then the probability that 0 persons will not have gotten back his or own card in any of those 15 rounds would be

I corrected a mistake in my work (I wrote a beta in the denominator instead of a gamma) when I substituted equ3b into equ4b, and I now have the same answer as bob bundy.

Because they are equal (just multiply in that negative one).  This is useful to do sometimes for algebraic simplification.

I solved for A, elliminating A*,e,r, treating all symbols as variables.  Note that I capitalized a to A to differentiate it between alpha a little more.  In fact, you will find from the work below that alpha does not even make up A at all.

I provided all steps I took so that you can double check my work (I don't solve systems of equations that often anymore).

Problem:
Solve for A in terms of anything but A*,e, and r using the following system of equations:

Solution:

EDIT: I corrected the mistake and got the correct answer.

I know it's almost been a year since this thread's last post, but I made an "Adjusted Pascal's Triangle" to create these power sum formulas for positive integers, and I finally got around to posting it on this forum.

Which I drew from the following formula I derived "from scratch":

(Putting in the n at the top of the triangle for sum of i^0 worked out perfectly, even though the formula itself cannot compute the correct value for R = 0).

Adjusting this triangle and using the method that knighthawk used in the previous post, I created an "Adjusted Pascal's Triangle" to compute the Bernoulli numbers which can be viewed here (it's too wide to post here, I think).

Basically, I found that

, and I used that to create that "Bernoulli Triangle".

I then wrote the following recursion formula from that Bernoulli triangle.

where

, where you can calculate a Bernoulli number in terms of its predecessor Bernoulli numbers.

Has anyone seen similar images like these "Adjusted Pascal Triangles" to compute the power sum and Bernoulli numbers visually?  I'm curious because I never heard of either, particularly the sum of power one, in high school or college math courses.

Oh.  That was actually the first thing that popped into my head, but it stayed in there for such a short time that I simply forgot.:)

I have sadly never heard of Abel's formula until today, but it definitely shows that the sum of products is not equal to the product of the sums.  Great point.  I'll have to take a closer look at the formula.

I think something that might throw off a lot of students is the fact that:

, but I could be wrong.

Oh, okay.  You're very welcome.  If there is any part of it which you need clarification, let me know...I didn't really write out any of the proof in English.:)

Ah, this wasn't too bad using algebra.

Q.E.D.

The only flawed step was when he divided by

Can you explain why it's against the rules?  If it means anything, (a/b)^(Log[a^3/b]/Log[a/b]) = a^3/b with Mathematica.

The 3D Graphs of the two are a little different though.  Is that what you are talking about?

Hi, anonimnystefy,

I ask kindly that you show your reasons (in fact, I asked in my last post for you to explain yourself).

The starting assumption was:

This equality is true whether a = b or a does not equal b.  As long as a and b are non-zero, it's a true statement.

And yes, there is a contradiction when he wrote

So, unless I have not fully identified what you found as incorrect yet, the only portion of his argument that was incorrect was that he didn't mention that a and b must be non-zero (which was pretty obvious to me, even though he didn't state it).

I'm not sure how he made any changes to the function in his work.  You have been stating that he has made an error by subtracting 1 from the numerator and the denominator when clearly he followed the properties of logs.

Here are all of the steps (flawless):

Of course, those steps can only be true if he would have specified that a and b are non zero in addition to the

His overall argument (if he would have put in those extra restrictions which avoids 0/0) reminds me of set theory.

This makes sense in a way why Don Blazys' argument doesn't work for a = 0 and thus b = 0, because the empty set does not have the possibility of being a proper subset of itself:

But then again, what about the next natural question that follows:

Is

Certainly this cannot be true, but what happens if we exclude the empty set from set theory?  The definition of a set would change...

So maybe that's where this argument might be headed...

Very interesting argument for non-zero a (and thus non-zero b), which is a big chuck of the reals if you ask me!

I know very little about set theory, so maybe someone could add on to this/point out some things I said which might be incorrect about set theory.