How many different color patterns are possible?

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The discussion revolves around calculating the number of different color patterns for planting 12 tulip bulbs in a 4x3 grid, consisting of 4 white, 4 red, and 4 yellow bulbs. For part a, the correct formula for the total arrangements is 12!/4!4!4!, confirming the initial calculation. In part b, treating each row as a single element leads to the conclusion that there are 3! arrangements for rows of the same color. For part c, the calculation involves considering the color in the corners, resulting in 3 x 8!/4!4! for the total patterns. The participants confirm the accuracy of these calculations and clarify the reasoning behind them.
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Homework Statement



A gardener has 12 tulip bulbs to plant in a box (4 columns by 3 rows). There are 4 white, 4 red and 4 yellow bulbs, and exactly one bulb is planted in each small square.
a. How many different color patterns are possible?
b. How many of these patterns have each row entirely the same color?
c. How many of these patterns have the four corner squares the same color?


Homework Equations





The Attempt at a Solution



a. I think this one is 12!/4!4!4!
b. and i think this one is 3!
Could I just get a confirmation of these 2?

c. Since the 4 bulbs of one color would be in the corner places, take 4 away from the original 12 spots available to get the other 2 colors' placement possibilities: 8!/4!4! I'm not sure if that's right and I think that I have to multiply something to represent the color that is in the 4 corners

Sorry I forgot to make a good title for the post...
 
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a. This is correct. For a problem a such, you first suppose that each ball is different. This would give you 12! possibilities. Now since you have 3 kinds of balls, you eliminate reoccurrences counter in 12!. This would give 12!/4!4!4!

b. How about you treat each row as a single element? How many elements would you have then? How many different arrangements?

c. This is correct for 1 color. Now, consider if the color in the four corners change. To what would the respective number of possibilities add up?
 
for b, isn't that 3!? And I have considered that for c, but I just don't know how to calculate that, would I multiply it by 3 or something?
 
You are correct on both points :smile:
 
so c is 3 X 8!/4!4! ?
 
Yes it is.
 
wow lol it was so simple...thanks a lot for helping me clear that up
 

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