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Suppose we want to place red, green and yellow balls into 7 distinguishable urns so that there is exactly one ball in each urn. In how many ways can this be done if we have an unlimited supply of balls of each colour, if balls of the same colour are indistinguishable, and if we want to use:

i) exactly 3 red balls;

ii) exactly 3 red and exactly 2 green balls;

iii) exactly 3 red and at least 2 green balls.

I'm unsure of how to proceed with this problem. At first, I thought there could be ${7 \choose 1}{6 \choose 1}{5 \choose 1} = 210$ choices of where to put 3 red balls, but I don't know if this is correct, or indeed how to work out the next two problems.

Please help!

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Suppose we want to place red, green and yellow balls into seven distinguishable urns so that there is exactly one ball in each urn. In how many ways can this be done if we have an unlimited supply of balls of each colour, if balls of the same colour are indistinguishable, and if we want to use exactly three red balls.

Choose which three of the seven urns will receive a red ball, which can be done in $\binom{7}{3}$ ways. For each of the four remaining urns, we have two choices. We can either place a green ball or a yellow ball in the urn. This can be done in $2^4$ ways. Hence, the number of ways the balls can be distributed is $$\binom{7}{3} \cdot 2^4$$

Suppose we want to place red, green and yellow balls into seven distinguishable urns so that there is exactly one ball in each urn. In how many ways can this be done if we have an unlimited supply of balls of each colour, if balls of the same colour are indistinguishable, and if we want to use exactly three red and exactly two green balls.

Choose which three of the seven urns will receive a red ball. Choose which two of the remaining four urns will receive a green ball. Fill the remaining two urns with a yellow ball.

$$\binom{7}{3}\binom{4}{2}\binom{2}{2}$$

Suppose we want to place red, green and yellow balls into seven distinguishable urns so that there is exactly one ball in each urn. In how many ways can this be done if we have an unlimited supply of balls of each colour, if balls of the same colour are indistinguishable, and if we want to use exactly three red and at least two green balls.

There are three cases:

  1. We use exactly three red, two green, and two yellow balls.
  2. We use exactly three red balls, three green balls, and one yellow ball.
  3. We use exactly three red and four green balls.

In each case, choose the positions of the three red balls, choose which of the four remaining urns will receive a green ball, then place a yellow ball in each remaining empty urn.

$$\binom{7}{3}\binom{4}{2}\binom{2}{2} + \binom{7}{3}\binom{4}{3}\binom{1}{1} + \binom{7}{3}\binom{4}{4}\binom{0}{0}$$

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To answer i), what we need is the amount of 4-tuples with greens and yellow multiplied by the amount of ways to position the 3 reds. The first factor is $2^4$ and the second is $\binom{7}{3}$.

As to ii), the first factor we need is the amount of permutations of $\{r,r,r,g,g\}$ which is $5!/3!/2!$. After that the remaining 2 balls must be yellow and they can be positioned in $\binom{7}{2}$ ways.

iii), This is just the sum of three cases (2, 3 or 4 greens). The first is known from ii). The other cases are similar to ii) with counts: $(6!/3! /3!)\binom{7}{1}$ and $(7!/3!/4!)\binom{7}{0}$

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Hint: for question (i), there are four cases:

  • 3 red balls, 4 green balls, 0 yellow balls
  • 3 red balls, 3 green balls, 1 yellow balls
  • 3 red balls, 2 green balls, 2 yellow balls
  • 3 red balls, 1 green balls, 3 yellow balls
  • 3 red balls, 0 green balls, 4 yellow balls

Can you count how many ways are there in each case?
Bonus Hint: for, say, the second case, it amounts to choosing 3 urns for the color red, 3 urns for the color green, and 1 urn for the color yellow. This is because the balls are indistinguishable.

For questions (ii) and (iii), the same reasoning applies.

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