I would appreciate some help with this

Within a box there are $6$ indistinguishable balls and $6$ different pairs of balls.

In how many way can be chosen $6$ balls from the box?

In how many ways can be distributed among three different people?

My attempt for the first question:

First I separated the balls into two classes: Those that are unpaired (Class A, that happen to be indistinguishable) and those who are paired (Class B). Then I began considering the cases where we have:

  • $0$ pairs from class B and $6$ from class A.
  • $1$ pair from class B and $4$ balls from class A.
  • $2$ pairs from B and $2$ from A.
  • $3$ pairs from B and $0$ from A.

According to me:

$$\binom{6}0\binom{6}6+\binom{6}1\binom{6}4+\binom{6}2\binom{6}2+\binom{6}3\binom{6}0 = 336$$

But the correct answer is $64$

My attempt for the 2nd question:

We can think on the people as three different boxes

This part can be separated into two. First we find the ways of distributing 6 indistinguishable balls in 3 different boxes. As fas as I understand this is reduced to a stars and bars problem. So we have


different ways.

For the second part we have $6$ different pairs of balls that have to be distribute in $3$ different boxes. First put the pairs in some arrange manner, labeled from $1$ to $6$. For the first there will be $3$ options. And the same is true for the rest of pairs. So we have $3^6$ ways of distributing this pairs among 3 different boxes. Then by the rule of sum we have $$\binom{6+3-1}3 + 3^6 = 785$$

But the textbook says that is $20412$

  • $\begingroup$ If I interpret your question as: "6 blue balls, 2 white balls, 2 black balls, 2 red balls" where two balls of similar color are indistinguishable from each other, ... does that change the meaning of your question? $\endgroup$
    – blackpen
    Nov 8, 2016 at 20:52
  • $\begingroup$ @blackpen Nope it doesn't change. That is a more concrete way to address the problem (: $\endgroup$
    – asd
    Nov 8, 2016 at 21:35
  • 1
    $\begingroup$ "1 pair from class B and 4 balls from class A.", ... Why are you selecting "pairs" of balls? Is that a requirement? $\endgroup$
    – blackpen
    Nov 8, 2016 at 23:22
  • $\begingroup$ @blackpen is one of the possible drawing from the box. It is supposed that each pair is tied together. $\endgroup$
    – asd
    Nov 9, 2016 at 6:08

2 Answers 2


As I interpret the question, the box contains $18$ balls: $\{6a, 2b, 2c, 2d, 2e, 2f, 2g\}$. I would undertake this by inclusion/exclusion as follows:

First, choosing $6$ balls from $7$ categories has ${12 \choose 6} = 924$ options. However this includes a lot of cases where there are more than $2$ balls chosen from the latter six categories. So calculating how many ways to choose as above but with $3$ or more balls from category $b$, say, gives ${9 \choose 6} = 84$ options for that for each pair category. We also need the cases which would break two restrictions at once; the cases where we would select $3$ balls from two of the pair categories, which can only be done ${6 \choose 2} = 15$ ways total.

Then combining this knowledge, we can strike the single-constraint breaking cases and add back in the double-counted double-constraint breaking cases:

$${12 \choose 6} - 6{9 \choose 6} + {6 \choose 2} = 924 - 6\cdot 84 +15 = 435$$

Of course this is not close to the book answer, or indeed your answer, but I leave it as an example of selection with constraints.


If you assume the balls as: aaaaaa, bb, cc, dd, ee, ff, gg (with the condition that any of bb, cc, dd, ee, ff, gg can only be selected as "take both or none").

As you said, there are four ways:

  • select 6 a's. Result = $\binom{6}{0}$=1
  • select 4 a's and rest others. Result = $\binom{6}{1}$=6
  • select 2 a's and rest others. Result = $\binom{6}{2}$=15
  • select 0 a's and rest others. Result = $\binom{6}{3}$=20

Total = 42.

For some reason, the expected answer 64 happens to be equal to $\binom{7}0+\binom{7}1+\binom{7}2+\binom{7}3$


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