A box contains two white, three black, and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw.

The answer given is $64$. I attempted the problem with two different approaches.

My two different attempts:

Attempt #1
Number of ways in which any $3$ balls can be drawn out of $9$ balls $$\binom{9}{3}=84$$ Number of ways in which no black balls are drawn out of $9$ balls (choosing $3$ balls from the remaining $6$ balls) $$\binom{6}{3}=20$$ Thus, the number of ways of choosing at least $1$ black ball is $$84-20=64$$

Attempt #2
Number of ways of drawing $1$ black ball out of $3$ black balls $$\binom{3}{1}=3$$ Now, we have to draw two more balls, we can choose those balls from the $8$ remaining balls $$\binom{8}{2}=28$$ Since both the above events are associated with each other, by fundamental principle of counting, the number of ways of drawing at least one black ball out of the $9$ balls is $$3\times28=84$$

I think my second attempt should also be right. Please explain what I'm doing wrong with my second attempt.


The second attempt counts the combination $\{B_1,B_2,B_3\}$ three times as $(B_1, \{B_2, B_3\})$, $(B_2, \{B_3, B_1\})$ and $(B_3, \{B_1, B_2\})$.

N.B: $(B_i, \{B_j, B_k\})$ means pick $B_i$ first, followed by the combination $\{B_j, B_k\}$.

We have the basic product rule

$$|A\times B| = \#\{(a,b) \mid A \times B\} = |A| \times |B|$$

for cardinals which holds for any sets $A$ and $B$.

By writing $3 \times 28$, you are actually counting $\{B_1, B_2, B_3\} \times \{B_2,B_3, \dots\}$, which doesn't answer the problem.

In , $(a,b)$ is defined as $\{a,\{a,b\}\}$, so order matters.

  • $\begingroup$ Would you please elaborate the answer? I get that the second attempt is wrong, but I'm unable to understand it clearly. $\endgroup$ – Azelf Dec 7 '18 at 20:14
  • $\begingroup$ @Azelf My original notation would be clear enough. I've edited my answer to make it even clearer. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 7 '18 at 20:57

Your first answer is correct.

A selection of three balls that includes at least one black ball has either one black ball and two of the other six balls, two blacks balls and one of the other six balls, or three black balls and none of the other six balls. Therefore, the number of ways of selecting at least one black ball when three balls are selected from two white, three black, and four red balls is $$\binom{3}{1}\binom{6}{2} + \binom{3}{2}\binom{6}{1} + \binom{3}{3}\binom{6}{0} = 45 + 18 + 1 = 64$$

Why was your second method incorrect?

You counted each case in which $k$ black balls were selected $k$ times, once for each of the $k$ ways you could have designated one of those black balls as the designated black ball. Notice that $$\color{red}{\binom{1}{1}}\binom{3}{1}\binom{6}{2} + \color{red}{\binom{2}{1}}\binom{3}{2}\binom{6}{1} + \color{red}{\binom{3}{1}}\binom{3}{3}\binom{6}{0} = 45 + 36 + 3 = 84$$ To illustrate, place numbers on the black balls. If you select black balls $b_1$ and $b_2$ and a red ball, you count this selection twice: \begin{array}{c c} \text{designated black ball} & \text{additional balls}\\ b_1 & b_2, r\\ b_2 & b_1, r \end{array} If you select all three black balls, you count this selection three times: \begin{array}{c c} \text{designated black ball} & \text{additional balls}\\ b_1 & b_2, b_3\\ b_2 & b_1, b_3\\ b_3 & b_1, b_2 \end{array}


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