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You have infinitely many balls and each of them is colored with one of the $C$ colors. You decided to fill each of the $N$ boxes $(B_1, B_2, B_3, \ldots, B_N)$ with exactly one ball. In how many ways can you do that? Two ways are considered different if there is at least one box in one way that has different colored ball than in the other way.

I think the answer is $C^N$. Am I correct?

What would happen if the condition exactly one is removed?

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    $\begingroup$ In effect you are counting the $N$-tuples of colors, so yes, you are correct. $\endgroup$
    – Brian M. Scott
    Jan 18, 2016 at 2:03

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Each box can be filled in $C$ ways as tgere are C different colour balls so $N$ boxes can be filled in $C.C.C....(N times)=C^{N}$ do i t for $2,3$ balls it will be more clear

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  • $\begingroup$ What would happen if the condition "exactly one" is removed $\endgroup$
    – Orion
    Jan 19, 2016 at 12:23
  • $\begingroup$ @ManishGaurav: If there is no limit to the number of balls per box, there are infinitely many solutions (provided $C>0$). $\endgroup$
    – Marc van Leeuwen
    Jan 19, 2016 at 12:28
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$C^N$ is correct.

If the condition 'exactly one' is removed, and you have an infinite number of balls, you have an infinite amount of possibilities to fill each box (put 1, put 2, put 3, ...), so the number of combinations is infinite.

If you meant by 'removing the condition' only 'removing the word exact', so you can effectively put 0 or 1 balls in each box, just consider 0 balls as a new 'color'. That gives you $(C+1)^N$ possibilities

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  • $\begingroup$ Thanks, this was very helpful. $\endgroup$
    – Orion
    Jan 21, 2016 at 1:58

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