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1.How many ways are there to put 6 balls in 3 boxes if the balls are not distinguishable but the boxes are?

 

2. How many ways are there to put 6 balls in 3 boxes if the balls are distinguishable but the boxes are not?

 

3. How many ways are there to put 6 balls in 3 boxes if three balls are indistinguishably white, three are indistinguishably black, and the boxes are distinguishable?

 Mar 17, 2019
 #1
avatar+4249 
+3

1. My favorite type of problem! Get ready for Stars and Bars! To read more about "Stars and Bars," check here: https://math.stackexchange.com/questions/910809/how-to-use-stars-and-bars

 

Anyway, when we plug \(n=6\) and \(k=3\)\(\), we have \(\binom{6+3-1}{3-1}=\binom{8}{2}=4*7=\boxed{28}.\)

 

2. We break this up into cases. To create 6, with three numbers, we can use: 6-0-0, 5-1-0, 4-2-0, 4-1-1, 3-3-0, 3-2-1, and. 2-2-2.

 

6-0-0: Only one way

5-1-0: Take the ball that is alone, so 6 ways.

4-2-0: 6C2=15 ways for the two balls in their own box.

4-1-1: Again 15 ways, pick two of the six balls, but they cannot go in the box of 4.

3-3-0: 5C2=10 ways, by selecting two of the remaining five balls.

3-2-1: 6*10=60 ways by picking any of the six balls to be by itself.

2-2-2: 15 ways? First, you think they are distinguishable, yet you overcount by a factor 3!=6.

 

Thus, adding them all up, we get \(1 + 6 + 15 + 15 + 10 + 60 + 15 = \boxed{122}. \)

 

 

3. Try this on your own! Hint: Just multiply by the same number of outcomes of the white balls for the black balls!

 Mar 17, 2019
 #2
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+1

Can you also solve this? How many ways are there to put 6 balls in 3 boxes if the balls are not distinguishable and neither are the boxes?

 Mar 17, 2019
 #3
avatar+4249 
+2

Sure. Here, we are dealing with indistinguishable balls and indistinguishable boxes. 

Now, we have to count the number of ways to separate the six balls in three boxes.

We can have 6-0-0, 5-1-0, 4-2-0, 4-1-1, 3-3-0, 3-2-1, and 2-2-2, so there are \(\boxed{7}\) ways.

 

smileysmiley

tertre  Mar 17, 2019
 #4
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+2

Thanks for your speedy reply! 

 

:)

 

smiley

Guest Mar 17, 2019
 #5
avatar+101161 
0

Thanks, tertre....these are always difficult   !!!

 

And

 

Nice pic, guest  !!!

 

 

cool cool cool

 Mar 17, 2019
 #6
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0

Thanks ChPill!!

 

I agree w/ you that these problems are not diificult.

Guest Mar 17, 2019
 #7
avatar+4249 
+2

Thank you, CPhill! smileysmiley

tertre  Mar 17, 2019

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