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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.
God, years since I did this, My guess would be 20, (5!/3!)
Fifteen.
I mean, fifteen?
No correct answers so far.
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?
You have to use all three colors, but you can use 1 color more than once, obviously otherwise it would be impossible :P
RRRBG
153 Just a random guess.
1800
The answer is 42
Nope :)Quote:
Originally Posted by xbones
Is it a trick question??
6 - would be my answer if it was
nevermind, didn't read the example -
158
I am assuming order matters.
As in red, red, blue, blue, green different from green, blue, blue, red, red
Is this true?
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
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.
I meant 148Quote:
Originally Posted by xbones
Heatman's method would give him the answer but he has missed something.
XBones Is close but no cigar ;)
I'm all in!
Thats my usual response when the math gets too hard! :lol:
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
Case 1 is incorrect ;)
147 ( 120+27 )= 147
Yeah - You are close, you have made a mistake and I know what it is as I done the same thing :)
Ah, 10 * 2 so is it 150?
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
very cool
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Doing a brute force count, it looks like 36...erm, I mean 243...um, I mean 120
Actually one of my early answers was 243, but that seemed like too many and I wasn't awake enough to commit to memory how I arrived at 120:
RRR--
R-RR-
R--RR
RR-R-
RR--R
R--RR
-RRR-
-RR-R
-R-RR
--RRR
that's 10 different positional combinations for any one color, meaning 30 (3*10) different positional combinations if we are given 3 and only 3 colors
for each positional combinations where 3 of one color is present, there are also 2 different positional combinations for each of the other colors
example:
GBRRR
BGRRR
RBGGG
BRGGG
RGBBB
GRBBB
that gives 60 total possibilities when 3 of one color is present
RR---
R-R--
R--R-
R---R
-RR--
-R-R-
-R--R
--RR-
--R-R
---RR
interesting...there are ALSO 10 possible positional combinations when only 2 positions are taken up by one color
That means there are 20 possible positional combinations for each combination of 2 colors, giving 60 (20*3 combinations of 2 colors[RB, RG, BG] holding 2 positions each)
That's 120 possible positional combinations when 2 or 3 colors is present
Tell me what I missed
I actually done a forced count first using 20,000 random numbers five in length between one and three. I then eliminated duplicates to arrive at 243 total combinations.