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Dr. Bob is Robert Hsiung, MD,
bob@dr-bob.orgShown: posts 1 to 15 of 15. This is the beginning of the thread.
Posted by Alexus on September 6, 2006, at 1:37:34
For the mathematically inclined.
If there is a dx category (BPD) with 9 symptoms...
And you need to have 5 or 6 or 7 or 8 or 9 of those symptoms to meet criteria for the dx...
Then how many different ways are there to meet the criteria for the dx?You could have...
1+2+3+4+5
or
1+2+3+4+5+6
or
2+3+4+5+6
or
1+3+5+7+9I'm sure you get the general idea.
But...
There must be a systematic way...(Someone computed 54 ways but I don't know how and I don't know whether that is true)
Posted by curtm on September 6, 2006, at 12:40:51
In reply to A math puzzle, posted by Alexus on September 6, 2006, at 1:37:34
Alexus, that's just mean. I love math puzzles. Now I will be up nights trying to figure this one out.
:)
Posted by Jost on September 6, 2006, at 13:50:34
In reply to A math puzzle, posted by Alexus on September 6, 2006, at 1:37:34
It's little complicated, and I have to try to get a nap--
but it's a variation on getting combinations of things, mathematically--
nCk = The number of combinations of n things taken k at a time--
the variation is because you don't want all the combinations of 9 things, but only those that have more than 5 things in them-- so the formula isn't the simple combination formula-=
you can google it-- combinations permutations--
but my math is of the musty, HS variety--
Jost
Posted by curtm on September 6, 2006, at 14:13:35
In reply to A math puzzle, posted by Alexus on September 6, 2006, at 1:37:34
Get ready and pay attention!
1 set of 9
(each set of 9 has 9 sets of 8) 1x9=9
(each set of 8 has 8 sets of 7) 9x8=72
(each set of 7 has 7 sets of 6) 8x72=441
(each set of 6 has 6 sets of 5) 6x441=2646Total: 3169
Posted by James K on September 6, 2006, at 14:28:32
In reply to I think I figured it out! » Alexus, posted by curtm on September 6, 2006, at 14:13:35
I hope your right, I was about to start typing them out, and I would have been pretty angry at myself about an hour into it.
I always solve things the most literal way I can think of.
James K
Posted by curtm on September 6, 2006, at 14:35:40
In reply to I think I figured it out! » Alexus, posted by curtm on September 6, 2006, at 14:13:35
> Get ready and pay attention!
>
> 1 set of 9
> (each set of 9 has 9 sets of 8) 1x9=9
> (each set of 8 has 8 sets of 7) 9x8=72
> (each set of 7 has 7 sets of 6) 8x72=441
> (each set of 6 has 6 sets of 5) **TYPO**x441=2646
>
> Total: 3169
1 set of 9
(each set of 9 has 9 sets of 8) 1x9=9
(each set of 8 has 8 sets of 7) 9x8=72
(each set of 7 has 7 sets of 6) 8x72=441
(each set of 6 has 6 sets of 5) 7x441=2646Total: 3169
Posted by Phillipa on September 6, 2006, at 15:26:42
In reply to Oops a typo! dammit » curtm, posted by curtm on September 6, 2006, at 14:35:40
So curtm is smart too? I never would have believed you'd figure it out. And at work no less. Love Phillipa
Posted by Jost on September 6, 2006, at 18:10:47
In reply to Re: Oops a typo! dammit » curtm, posted by Phillipa on September 6, 2006, at 15:26:42
Okay, I get 126-- but my math is er um not so great.
did you say 54?
hmmmm.. (but my math is uh pretty bad...)
Jost
Posted by finelinebob on September 6, 2006, at 21:36:53
In reply to A math puzzle, posted by Alexus on September 6, 2006, at 1:37:34
There is a systematic way. It's called a Combination. Cominations can either allow for repetition or not, but since it doesn't matter what order the dx's are presented we'll do it without.
(in other words, if it mattered that you were BP I THEN ADD, and that was different than ADD THEN BP I, we'd need to include repetition).
General formula is n!/[k!(n-k)!. If you have 9 dx's and you need to find out how many combinations there are if you take them 5 at a time, then n=9, k=5, and the result would be 9!/[5! (9-5)!]
If you don't know what that "!" means, it's a factorial. 3! = 3*2*1 = 6. 0! is defined as 1.
Your problem is that you have more than one combination. You have these:
n=9, k=5
n=9, k=6
n=9, k=7
n=9, k=8
n=9, k=9
Add them all together and you get your answer.Oh. That would be 256.
http://home.ubalt.edu/ntsbarsh/Business-stat/otherapplets/ComCount.htm#rjava3
flb
Posted by Alexus on September 7, 2006, at 1:31:17
In reply to Re: I think I figured it out! » curtm, posted by James K on September 6, 2006, at 14:28:32
> I hope your right, I was about to start typing them out, and I would have been pretty angry at myself about an hour into it.
> I always solve things the most literal way I can think of.Me too :-)
I was thinking of doing a truth table (that lists all the possibilities). Like this:1 2
P P
P A
A P
A A1= symptom 1
2= symptom 2
P= symptom is present
A= symptom is absentbut that is 4 rows (possibilities) for 2 symptoms.
with three symptoms it looks like this:1 2 3
P P P
P P A
P A P
P A A
A P P
A P A
A A P
A P Aso that is 8 rows for three variables.
so there would be 16 rows for four variables (is that right?)
and 32 rows for 5
and 64 rows for 6
and 128 rows for 7
and 256 rows for 8
and 512 rows for 9
(if i haven't stuffed up the addition)
i don't know if that is right...someone thought there would be 2 to the power of 9 rows on the truth table for 9 variables but i don't know what that is and if thats right...
then you would cross out the rows that have less than 5 P's on them. and what you are left with would be the answer :-)
but i don't have a big enough piece of paper (and i'm not sure on how big that piece of paper needs to be)
ak.
Posted by Alexus on September 7, 2006, at 1:33:06
In reply to Oops a typo! dammit » curtm, posted by curtm on September 6, 2006, at 14:35:40
> 1 set of 9
> (each set of 9 has 9 sets of 8) 1x9=9
> (each set of 8 has 8 sets of 7) 9x8=72
> (each set of 7 has 7 sets of 6) 8x72=441
> (each set of 6 has 6 sets of 5) 7x441=2646
> Total: 3169holy sh*t. i think i understand what you are doing...
i have no idea about that though...
:-)
Posted by Alexus on September 7, 2006, at 1:33:52
In reply to Re: Oops a typo! dammit, posted by Jost on September 6, 2006, at 18:10:47
> Okay, I get 126-- but my math is er um not so great.
> did you say 54?
> hmmmm.. (but my math is uh pretty bad...)hmm.
i guess i'll take bets...
:-)
Posted by Alexus on September 7, 2006, at 1:37:19
In reply to Re: A math puzzle, posted by finelinebob on September 6, 2006, at 21:36:53
> There is a systematic way. It's called a Combination. Cominations can either allow for repetition or not, but since it doesn't matter what order the dx's are presented we'll do it without.
okely dokely
> (in other words, if it mattered that you were BP I THEN ADD, and that was different than ADD THEN BP I, we'd need to include repetition).order of presentation of symptoms doesn't matter. just which are present or absent. i think i'm following you...
> General formula is n!/[k!(n-k)!.!
> If you have 9 dx's and you need to find out how many combinations there are if you take them 5 at a time, then n=9, k=5, and the result would be 9!/[5! (9-5)!]
!
(i'm a lot lost)
> If you don't know what that "!" means, it's a factorial. 3! = 3*2*1 = 6. 0! is defined as 1.okayyyyyyyyyy.
> Your problem is that you have more than one combination. You have these:
> n=9, k=5
> n=9, k=6
> n=9, k=7
> n=9, k=8
> n=9, k=9
> Add them all together and you get your answer.
> Oh. That would be 256.okay.
i'm really very stupid when it comes to math...thanks very much.
!
;-)
Posted by Alexus on September 7, 2006, at 6:44:11
In reply to Re: A math puzzle, posted by finelinebob on September 6, 2006, at 21:36:53
AAAHHHHHHHHHHH
I followed the link and tried to nut it out.
Messed up a lot.
>If you have 9 dx's and you need to find out how many combinations there are if you take them 5 at a time, then n=9, k=5Oops.
I was thinking n=2 (two possible truth values P and A) and k=9 (nine symptoms). Yeech. When I did stats for psychology they used to laugh at me because they said they never saw anyone make the errors I make. Its because of logic you see. They would give me things like 2X4X8X9X4 and I said I couldn't do it until they put the brackets in. It isn't a wff (well formed formula) without brackets, you see. But it goes like this, eh?:
{[(2X4)X8]X9}X4 (you do the stuff in the brackets first). Crap crap crap i'm crap at math :-(> Your problem is that you have more than one combination.
Ah, I see. I figured it would just tell me how many rows on the truth table... I didn't know how to go about getting rid of the rows with less than 5 P's...
You have these:
> n=9, k=5
> n=9, k=6
> n=9, k=7
> n=9, k=8
> n=9, k=9
> Add them all together and you get your answer.
> Oh. That would be 256.That makes sense (kinda semi sorta)
Thanks.:-)
Posted by curtm on September 7, 2006, at 9:22:58
In reply to Re: A math puzzle, posted by finelinebob on September 6, 2006, at 21:36:53
This is the end of the thread.
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