I have a question in my textbook.

A box contains $2$ white balls, $3$ black balls and $4$ red balls. In how many ways can $3$ balls be drawn from the box, if at least one black ball is to be included in the draw?

My assumption was that the ways to draw the fixed 'at least $1$ black ball' were same as the ways to draw $2$ or $3$ black balls, i.e. $1$. So, for each case;

$1$ black ball, and $2$ others, which is $1\times\binom62$

$2$ black ball, and $1$ other, which is $1\times\binom61$

$3$ black ball, and $0$ others, which is $1\times\binom60$

For a total of $22$ cases. But the solution in my textbook assumes otherwise. That each black ball is distinct. Is there something I missed in the statement or it is ambiguous and correct both ways?

  • $\begingroup$ I'd say the problem was poorly phrased. Assuming that they are, eventually, going to ask about probabilities then it would be clear that they meant to regard each ball as "distinct", so that $B_1,W_1, R_1$ is not the same as $B_2, W_1, R_1$. $\endgroup$
    – lulu
    Apr 14 '20 at 14:04
  • $\begingroup$ @lulu This is the end of the question. However, I agree with you that it is indeed poorly phrased. $\endgroup$
    – Rew
    Apr 14 '20 at 14:05
  • $\begingroup$ This ambiguity comes up a lot. How many way ways are there to throw a $3$ as the sum of two dice? You could say "only one since $\{1,2\}$ is the same as $\{2,1\}$. However, if you thinking of probability, it is much better to say that these are two different ways. Otherwise you have to deal with the fact that $\{1,2\}$ is twice as probable as $\{1,1\}$. $\endgroup$
    – lulu
    Apr 14 '20 at 14:10
  • 3
    $\begingroup$ Your solution lacks consequence. You assume that white balls are distinguishable, as well as red ones, but then you assume black ones are not. $\endgroup$ Apr 14 '20 at 14:10
  • $\begingroup$ How many ways to draw 3 black balls are there according to the textbook? $\endgroup$
    – user
    Apr 14 '20 at 14:53

This is a pretty standard way to ask for the number of $3$-element subsets of the set of $9$ balls that contain at least one black ball. I’d be somewhat surprised if the intended answer were not $\binom93-\binom63=84-20=64$ unless the larger context (e.g., text material that immediately precedes the question) clearly suggested either that the balls are distinguishable only by color or that the balls are to be drawn one at a time, and different orders are to be counted as distinct draws.

In any case, if you take the balls to be indistinguishable apart from color, there are only $6$ distinguishable outcomes with at least one black ball if order is not taken into account, namely, $BWW$, $BWR$, $BRR$, $BBW$, $BBR$, and $BBB$, and $19$ if order is taken into account. As Michal Adamaszek said, your answer is inconsistent, since you treat the black balls as indistinguishable but the white and red balls as distinguishable.


Maybe you can consider some balls of the same color as distinct to obtain the solution, but it is neither necessary nor helpful. The most efficient way is to use the complementary event without making distinction.

Number of ways to draw 3 balls from the box with at least black ball=

Number of ways to draw 3 balls from the box - Number of ways to draw 3 balls from the box with no black ball.

The drawings are 3 balls from 9 minus 3 red balls minus 1 white and 2 red balls minus 2 white and 1 red balls.

$$\{\text{3 balls}\}-\{\text{0w,3r}\}-\{\text{1w,2r}\}-\{\text{2w,1r}\}$$

$$\binom{9}{3}-\binom{2}{0}\cdot \binom{4}{3} -\binom{2}{1}\cdot \binom{4}{2}-\binom{2}{2}\cdot \binom{3}{1}$$



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