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mathsbrain

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Q: find the number of ways that 6 different coloured balls
can be placed in 3 non-identical urns so that no urn is empty.

Answer 1: 540

Answer 2: 1080

Which one is correct and why?
 
GoldyOrNugget

GoldyOrNugget

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When uncertain, bash out the cases!


If we put ONE ball in the first urn:
6C1 possibilities for this ball * (5C1 + 5C2 + 5C3 + 5C4), for each possible amount in the second urn

TWO balls in the first urn:

6C2 * (4C1 + 4C2 + 4C3)

THREE balls:

6C3 * (3C1 + 3C2)

FOUR balls:

6C4 * (2C1)

Add them all together, giving 540.
 
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mathsbrain

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My way of thinking is there are 3 cases:

Case 1: one urn has 4 balls, one urn has 1, another urn has 1.

So we do 6C4 * 2C1 * 1C1 to choose the balls, then arrange them so 3!, and we DO NOT divide by 2! because the urns are NOT IDENTICAL, is this correct? This is the part that i am confused, to divide by 2! or not. I think if the urns are identical then we divide.
 
GoldyOrNugget

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You still divide by 2! because (4,1,1) is the same as (4,1,1).
 
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mathsbrain

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so for example,

Case 1: balls B,C,D,E in urn A, ball A in urn B and ball F in urn C
Case 2: balls B,C,D,E in urn A, ball F in urn B and ball A in urn C

Would you call that same outcome or different? I would say, they are same outcome IF urns are identical, but this is clearly not the case right? can carrotstick help?
 
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jyu

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3(6c4)(2c1)(1c1)+3!(6c3)(3c2)(1c1)+1(6c2)(4c2)(2c2)=540
 
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