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Discussion Starter #1
Consider The Following:

We all know that our great cars were produced in one of three available colors. Now, if someone is lucky enough to own three Bullitts, he or she could thus have them in any of the following ten color combinations:

All green.
2 green and 1 black.
2 green and 1 blue.
All black.
2 black and 1 green.
2 black and 1 blue.
All blue.
2 blue and 1 green.
2 blue and 1 black.
1 green, 1 black and 1 blue.

NOW...here is the tough part: when figuring that any one of the cars listed above could, or could not, have the Mach stereo option how many combinations - encompassing BOTH color and the Mach stereo option - are now possible, e.g., three green ones all of which have the mach, three green ones with two machs and one non-mach, etc., etc., etc. There is a relatively easy way to figure this question out without having to really do much calculating at all...do you see it?

Good Luck!
 

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Sorry, Stan, but my brain is still on vacation. Don't leave me in suspense too long.
 

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Multiply everything times 2?
So 20 possible combos? (with this situation)

I don't know, I was a Geology major.

_________________
DHG #3875
Modifications:
BULLITT chase scene recreated on rear dash
with Revell 1/25 scale 68 GT 390 Fastback and
68 Charger R/T metal body models. Unfortunately
the hub caps don't come off the Charger.



<font size=-1>[ This Message was edited by: PY on 2001-12-16 13:02 ]</font>
 

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No wait! Multiply each of the 10 categories
by 2, so 60 combos!

<font size=-1>[ This Message was edited by: PY on 2001-12-16 14:46 ]</font>
 

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I think there are 8 MACH/no Mach combinations per group of 3 cars. Assuming that each car of the group can or cannot have a Mach. Therefore, I think the combinations are 80. There should be simple equation to figure this out, I can't remember it though.
 

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C(N,R)=P(N,R)/P(R,R)

C(6,3)=P(6,3)/P(3,3) = (6*5*4)/(3*2*1)=

(6*5*4)/6 = 5*4 = 20 :-B

Or like some one said, "twice as many".

<font size=-1>[ This Message was edited by: SteveBullitt on 2001-12-16 13:39 ]</font>
 

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I got it! One.

Isn't Dark Highland Green with the Mach 460 the ONLY combination that counts!

HeHe.

Actually, the answer is 56.
 

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<TABLE BORDER=0 ALIGN=CENTER WIDTH=85%><TR><TD><font size=-1>Quote:</font><HR></TD></TR><TR><TD><FONT SIZE=-1><BLOCKQUOTE>
On 2001-12-16 13:50, 01GTCOUPE wrote:
I got it! One.

Isn't Dark Highland Green with the Mach 460 the ONLY combination that counts!

HeHe.

Actually, the answer is 56.

GOOD ANSWER!!!

</BLOCKQUOTE></FONT></TD></TR><TR><TD><HR></TD></TR></TABLE>
 

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<TABLE BORDER=0 ALIGN=CENTER WIDTH=85%><TR><TD><font size=-1>Quote:</font><HR></TD></TR><TR><TD><FONT SIZE=-1><BLOCKQUOTE>
On 2001-12-16 15:12, Jimmy Ray wrote:
I need another beer????????????????????????????????????????

</BLOCKQUOTE></FONT></TD></TR><TR><TD><HR></TD></TR></TABLE>

I'm right with you on this one, Jimmy.

 

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Damn! 01GTCOUPE got it!
I think 56 DIFFERENT COMBOS
is correct.
I was counting CARS not
COMBOS before. Kind of like apples
and oranges.

I don't know, my degree is in Geology.

BEER PLEASE!

DHG #3875
Modifications:
BULLITT chase scene recreated on rear dash
with Revell 1/25 scale 68 GT 390 Fastback and
68 Charger R/T metal body models. Unfortunately
the hub caps don't come off the Charger.

<font size=-1>[ This Message was edited by: PY on 2001-12-16 15:38 ]</font>
 

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OK, I'm working on it now. When all 3 are the same color, there are 4 possible combos: all with radios, none with radios, 2 with radios or one with radio. Once for each color gives you 12 combo.

When 2 are 1 color and 1 another, there are 6 possible combos: all with, all without, same color 2 with, 2 of different color with, or 1 with (twice because there's 2 colors involved.) That's 36 more combo's

Last there's one of each color. Again, they can be: all with, all without, 2 with and 1 without (3 color combo's) or 1 with and 2 without (3 color combo's) That's 8 more combo's.

36+12+8=56 combos.
 

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Hey Bullittgirl, I think you are right.
I had to sit down with a sheet of paper
just to keep everything straight. Eventually
came up with 56 also but 01GTCOUPE beat us
to it. Who knows, maybe we are all wrong!
HA!
 

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Okay, I blew it in my first attempt because I forgot that you could have more than one of the same kind.

There are six possible cars:
A = GRN/460
B = GRN
C = BLK/460
D = BLK
E = BLU/460
F = BLU

Here are the possible combos:
AAA AAB AAC AAD AAE AAF
ABB ABC ABD ABE ABF
ACC ACD ACE ACF
ADD ADE ADF
AEE AEF
AFF

BBB BBC BBD BBE BBF
BCC BCD BCE BCF
BDD BDE BDF
BEE BEF
BFF

CCC CCD CCE CCF
CDD CDE CDF
CEE CEF
CFF

DDD DDE DDF
DEE DEF
DFF

EEE EEF
EFF

FFF

Or a total of 56 combos.
 

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Great!!! I have just polished off no. 9, but if weak memory serves, this was a simple matter of probability and statistics. Punch in the right numbers on the correct calculator, and you can even figure out your chances of winning the lottery. Too bad i'm feeling too good right now!
 

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Discussion Starter #18
Wow...now I am even confused! Let me check my math over dinner tonight and see how my answer compares to the other guesses! For all I know my original guess is wrong!

Happy Monday!
 

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Ya know, there's a name for this kind of problem, and a horrifyingly easy solution that doesn't involve drawing out all of the answers. Heck if I remember what it is, though. :roll:

----------20 minutes later------------

Permutation! That's it!

Easy solution: There are 6 possible values. The total number of permutations, or combinations, is the same as 6 * 5 * 4 * 3 * 2 * 1, or 6!, which is 720. BUT, in this problem, only 3 permutations can be chosen at a time. Take the total number of values (6), subtract the number of permutations to select at a time (3), take it's factorial (6-3)! = 6, and divide the total number of combinations by it. 720/6 = 120. Now, half of those are really just duplicates of another list, so divide that by 2, and you get 60. PY was right!

Don't believe me? Check this out. You'll have to do the last part (divide by 2) on your own.

For this problem, pick the type for "All k-permutations..." and the values for this problem are n=6 for 6 different values, and k=3 for only allowing 3 at a time. The coolest thing is the output. Sure, you can get a boring list of all the combinations, but it's much more fun to see the chess-board representations of all 120. :smile:


<font size=-1>[ This Message was edited by: Micah on 2001-12-17 15:53 ]</font>
 
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