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Problems:
1. How many ways 6 balls can be distributed among three boxes?
2. How many equations of the form $ax^2+bx+c=0$ can be formed if the coefficients are determined by throwing an ordinary six faced die.

I have confusion with the second problem to distinguish from the first problem.

Here, in the problem 1. Since each ball can be put into any one of the three boxes. So, the total number of ways in which $6$ balls can be put into three boxes is $3^6.$

and for the problem 2.
Why we can't think as: out of six numbers (1,2,3,4,5,6) each be taken by one of the three coefficients $a,b,c$. So the total number of ways should be $3^6$.
But the actual answer is $6^3$. Why? please

Primo
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2 Answers2

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For problem 2, you have 6 options for a, 6 for b, 6 for c, so $6^3$ is correct.

For problem 1, the question is whether the balls are distinguishable.

If so, then you have 3 options for ball 1, .., three for ball 6, so $3^6$ is correct.

If not, then you have to do https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

Pieter21
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You can choose any number in $(1,2,3,4,5,6)$ for the coefficient $a $ so there are 6 possible choices.

For coefficient $b$ you have 6 choices as well, so you have $6\cdot 6$ total up until now.

For coefficient $c $ you have another 6 choices so $6\cdot 6\cdot 6 = 6^3$ total ways.

RGS
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