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Toasty's Math Puzzel 1

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Toasty
Post Posted: Wed, 30 Jun 2004, 5:51am    Post subject: Toasty's Math Puzzel 1 Reply with quote
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I thought I'd liven up the forum a little with some off topic stuff.

Scenario:
There are five fence pannels and three pots of paint. One Blue, one Red and one Green. Every panel needs to be painted.

Example

Blue Blue Red Red Green

How many ways can you paint the fence panels using all three colors?

There is more than enough paint.


Last edited by Toasty on Wed, 30 Jun 2004, 6:37am; edited 1 time in total
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xbones
Post Posted: Wed, 30 Jun 2004, 6:14am    Post subject: Reply with quote
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God, years since I did this, My guess would be 20, (5!/3!)
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heatman
Post Posted: Wed, 30 Jun 2004, 6:33am    Post subject: Reply with quote
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Fifteen.
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heatman
Post Posted: Wed, 30 Jun 2004, 6:34am    Post subject: Reply with quote
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I mean, fifteen?
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Toasty
Post Posted: Wed, 30 Jun 2004, 6:36am    Post subject: Reply with quote
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No correct answers so far.
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heatman
Post Posted: Wed, 30 Jun 2004, 6:38am    Post subject: Reply with quote
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Point of clarification...

Is it okay to have all panels the same color, or do you mean you have to use all three colors? i.e. is Blue, Blue, Blue, Blue, Blue an option?
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Toasty
Post Posted: Wed, 30 Jun 2004, 6:57am    Post subject: Reply with quote
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You have to use all three colors, but you can use 1 color more than once, obviously otherwise it would be impossible Razz

RRRBG
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xbones
Post Posted: Wed, 30 Jun 2004, 7:02am    Post subject: Reply with quote
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153 Just a random guess.
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jmrogers7
Post Posted: Wed, 30 Jun 2004, 7:03am    Post subject: Reply with quote
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1800
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xbones
Post Posted: Wed, 30 Jun 2004, 7:05am    Post subject: Reply with quote
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The answer is 42
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Toasty
Post Posted: Wed, 30 Jun 2004, 7:08am    Post subject: Reply with quote
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xbones wrote:
The answer is 42


Nope Smile
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Yeah
Post Posted: Wed, 30 Jun 2004, 7:18am    Post subject: Reply with quote
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Is it a trick question??

6 - would be my answer if it was
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Yeah
Post Posted: Wed, 30 Jun 2004, 7:20am    Post subject: Reply with quote
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nevermind, didn't read the example -


Last edited by Yeah on Wed, 30 Jun 2004, 7:35am; edited 1 time in total
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xbones
Post Posted: Wed, 30 Jun 2004, 7:31am    Post subject: Reply with quote
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158
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Humphrind
Post Posted: Wed, 30 Jun 2004, 7:39am    Post subject: Reply with quote
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I am assuming order matters.

As in red, red, blue, blue, green different from green, blue, blue, red, red

Is this true?
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Toasty
Post Posted: Wed, 30 Jun 2004, 7:45am    Post subject: Reply with quote
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No one has it yet but someone has come close.

Its visually based, as long as another isn't visually the same and uses 3 colors it counts, i.e. RRRBG - GBRRR - RRRBG - GGGBR
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heatman
Post Posted: Wed, 30 Jun 2004, 7:50am    Post subject: Reply with quote
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So, if you didn't have to use all three colors, I think the answer is 243.

You've got three choices for each panel. Since order matters, these are independent choices, ie you've got three choices for panel 1, then for each color you pick for panel 1, you've got three choices for panel 2, or nine different combinations for two panels, then for each one of these you've got three choices for panel 3, etc. etc. etc. Thats 3X3X3X3X3 = 3^5 = 243

Then, you have to figure out how many 2 color and one color combinations there are. There are 3 one color combinations. That means we're down to 240.

Using the same reasoning above, I'd guess there are 2^5 = 32 two color comos. So my new, and final guess is 208.
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xbones
Post Posted: Wed, 30 Jun 2004, 7:54am    Post subject: Reply with quote
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xbones wrote:
158


I meant 148
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Toasty
Post Posted: Wed, 30 Jun 2004, 7:58am    Post subject: Reply with quote
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Heatman's method would give him the answer but he has missed something.

XBones Is close but no cigar Wink
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heatman
Post Posted: Wed, 30 Jun 2004, 8:02am    Post subject: Reply with quote
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I'm all in!
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heatman
Post Posted: Wed, 30 Jun 2004, 8:03am    Post subject: Reply with quote
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Thats my usual response when the math gets too hard! Laughing
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xbones
Post Posted: Wed, 30 Jun 2004, 8:06am    Post subject: Reply with quote
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My reasoning is:

I can have 1, 2 or 3 Reds.

Case 1 : 3 reds; there are 9 diff ways 3 reds can be on the fences, and only (2^2)-2 = 2 ways the rest can be Chosen, hence 9*2= 18

Case 2 : 2 Reds 10 diff ways , and (2^3)- 2 = 6 hence 10*6 = 60

Case 3 : 1 red Can be in 5 places, and (2^4)-2 = 14 hence 5*14 = 70
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Toasty
Post Posted: Wed, 30 Jun 2004, 8:26am    Post subject: Reply with quote
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Case 1 is incorrect Wink
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Yeah
Post Posted: Wed, 30 Jun 2004, 8:27am    Post subject: Reply with quote
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147 ( 120+27 )= 147
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Toasty
Post Posted: Wed, 30 Jun 2004, 8:34am    Post subject: Reply with quote
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Yeah - You are close, you have made a mistake and I know what it is as I done the same thing Smile
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xbones
Post Posted: Wed, 30 Jun 2004, 8:37am    Post subject: Reply with quote
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Ah, 10 * 2 so is it 150?
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Toasty
Post Posted: Wed, 30 Jun 2004, 8:50am    Post subject: Reply with quote
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and we have a winner!!!

We have 3 colors and 5 panels.

Step 1

Total combinations therefore are 3^5=243

Step 2

We now need to remove all the combinations where only 2 or less are used.

2^5=32 We have 3 combinations of paint Red/Blue Red Green Blue/Green

(2^5)*3=96

243-96=147

Step 3

We now have 147 however in step two we removed all red/blue/green combinations twice. i.e. once for Red/Blue and once for Red/Green which would remove RRRRR twice.

We add this 3 back to arrive at 150

Answer = 150


Last edited by Toasty on Thu, 01 Jul 2004, 12:16am; edited 1 time in total
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vasumi
Post Posted: Wed, 30 Jun 2004, 10:53pm    Post subject: Reply with quote
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very cool
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