Is this cool with you?

gAr · 2011-07-01 21:57:37

Hi,

Here's a python script.

cnt = 0
cnt10 = 0
for a in (1,2,4,8,16):
    for b in (1,2,4,8,16):
        for c in (1,2,4,8,16):
            for d in (1,2,4,8,16):
                for e in (1,2,4,8,16):
                    if a<=b and b<=c and c<=d and d<=e:
                        cnt += 1
                        print a,b,c,d,e
                        if a+b+c+d+e == 10:
                            cnt10 += 1
print cnt,cnt10

Correct, isn't it?

bobbym · 2011-07-01 22:07:45

Hi gAr;

I believe so. I am now getting 126 also. I do not where the error came from. That clears up one problem, thanks!

gAr · 2011-07-01 22:13:39

Hi bobbym,

Okay, nice to hear that!
Is 126 the answer then?

bobbym · 2011-07-01 22:18:59

Yes! Also there seems to be a simple combinatorial answer!

gAr · 2011-07-01 22:29:33

Cool!

there seems to be a simple combinatorial answer!

Number of terms in the expansion of (a+b+c+d+e)^5 ?

But I'm still a little confused.
Instead, if B had got (2,2,2,2,2) twice or (4,2,2,1,1) twice, will it still be the same answer?
Are the events of getting (2,2,2,2,2) and getting (4,2,2,1,1) equiprobable?

bobbym · 2011-07-01 22:38:02

Hi;

Those are equiprobable. Each of the 5 areas that he can hit all are equiprobable. Each one has a point value associated with it. So with one dart scoring 2 is just as likely as scoring 1. So with 5 darts 1,1,1,1,1 is as likely as 2,2,2,2,2 or 4,2,2,1,1.

gAr · 2011-07-01 22:46:53

Hi,

Okay, thanks.

If B had got only one game with a score of 10 (say 2,2,2,2,2), can we still estimate the number of games?

bobbym · 2011-07-01 22:58:22

Hi gAr;

The equiprobable aspect of the question was only brought in to state that there was no bias. The problem could have phrased in a way that removed any mention of a probability but I wanted to save that for later. When I knew a little more.

Anyways the number of games is:

As long as areas to hit equals darts. In this case.

gAr · 2011-07-01 23:17:37

Hi bobbym,

Okay. Thanks for the formula!
So they are the number of coefficients in the expansion of multinomial, like I doubted in #732.

bobbym · 2011-07-01 23:58:15

Hi gAr;

I am not following you there. The largest coefficient is 120 in that expansion. I am not sure how to get 126 out of it.

gAr · 2011-07-02 00:04:20

Hi,

I meant this: http://en.wikipedia.org/wiki/Multinomia … efficients

In our case:
m=5, n=5

bobbym · 2011-07-02 00:09:30

Hi gAr;

It will not work for 6 darts and 6 areas.

gAr · 2011-07-02 00:18:23

Hi,

I got it correct.

bobbym · 2011-07-02 00:26:40

Hi gAr;

I am getting

gAr · 2011-07-02 00:32:20

Hi bobbym,

You didn't include the 6th point value, did you??

bobbym · 2011-07-02 00:34:05

I used the same areas but just 6 darts. That is no good?

gAr · 2011-07-02 00:42:48

You told it wouldn't work for 6 darts and 6 areas.
Did you mean 6 darts and 5 areas?
For that I'm getting same as yours.

n is the number of darts
m is the number of areas.
Did you also assume the same way?

bobbym · 2011-07-02 00:44:58

Hi gAr;

The confusion is because the problem is stated in a way that is the same but is difficult to follow. Give me a second and I will phrase it more accurately.

The real question:

If I have the multiset
{1,1,1,1,1,2,2,2,2,2,4,4,4,4,4,8,8,8,8,8,16,16,16,16,16}

Using 5 of them how many restricted partitions ( order does not count ) can I form?

gAr · 2011-07-02 01:03:52

Yes, our answers match.

You got this for 6 darts and 5 areas?

bobbym · 2011-07-02 01:09:36

Hi gAr;

Yes, the new problem is;

{1,1,1,1,1,1,2,2,2,2,2,2,4,4,4,4,4,4,8,8,8,8,8,8,16,16,16,16,16,16}

The area really means 5 different types of numbers {1,2,4,8,16}. The 5 types stay the same, the number of them goes up.

Same as yours.

gAr · 2011-07-02 01:14:54

Hi,

Okay, so you thought of multisets, and I thought of multinomials. Nice!

bobbym · 2011-07-02 01:19:02

Hi gAr;

When I worked on this for someone else he used pegs on a board. I saw the connection with the restricted partition problem. Not much has been done on them. Everybody knows about partition numbers but when the set is finite then not so much is known.

This is the first improvement I have seen on this type in a while.

gAr · 2011-07-02 01:56:35

Hi bobbym,

Okay.
I haven't given multisets much thought.

bobbym · 2011-07-02 02:20:05

I have played with them a little. It is harder to find answers to particular ones rather than all of them. The generating function is harder.

gAr · 2011-07-02 02:35:20

Yeah, I'm still trying for a g.f.

Math Is Fun Forum

#726 2011-07-01 21:57:37

Re: Is this cool with you?

#727 2011-07-01 22:07:45

Re: Is this cool with you?

#728 2011-07-01 22:13:39

Re: Is this cool with you?

#729 2011-07-01 22:18:59

Re: Is this cool with you?

#730 2011-07-01 22:29:33

Re: Is this cool with you?

#731 2011-07-01 22:38:02

Re: Is this cool with you?

#732 2011-07-01 22:46:53

Re: Is this cool with you?

#733 2011-07-01 22:58:22

Re: Is this cool with you?

#734 2011-07-01 23:17:37

Re: Is this cool with you?

#735 2011-07-01 23:58:15

Re: Is this cool with you?

#736 2011-07-02 00:04:20

Re: Is this cool with you?

#737 2011-07-02 00:09:30

Re: Is this cool with you?

#738 2011-07-02 00:18:23

Re: Is this cool with you?

#739 2011-07-02 00:26:40

Re: Is this cool with you?

#740 2011-07-02 00:32:20

Re: Is this cool with you?

#741 2011-07-02 00:34:05

Re: Is this cool with you?

#742 2011-07-02 00:42:48

Re: Is this cool with you?

#743 2011-07-02 00:44:58

Re: Is this cool with you?

#744 2011-07-02 01:03:52

Re: Is this cool with you?

#745 2011-07-02 01:09:36

Re: Is this cool with you?

#746 2011-07-02 01:14:54

Re: Is this cool with you?

#747 2011-07-02 01:19:02

Re: Is this cool with you?

#748 2011-07-02 01:56:35

Re: Is this cool with you?

#749 2011-07-02 02:20:05

Re: Is this cool with you?

#750 2011-07-02 02:35:20

Re: Is this cool with you?

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