Question

A box contains two white balls, three black balls and four red balls. Balls of the same color are distinct. The number of ways in which three balls can be drawn from the box if at least one black ball is to be included in the draw is:
A. 32
B. 64
C. 128
D. None of these

Answer 1

Hint: It is given that a box contains 2 white balls, 3 black balls and 4 red balls and balls of the same color are distinct. Also given that three balls are drawn from the box and at least one black ball is to be included in the draw. We can break the question into three cases which contain 1 black ball, 2 black balls and 3 black balls.

Complete step by step solution: A box has 2W, 3B, 4R balls. We are using ‘B’ for black, ‘W’ for white and ‘R‘ for red.
We divide the above question into 3 cases and finally add them all.
Step 1: First we take the case in which one black ball is always included.
Case 1: If we get one black ball then the rest are either red or white.
(BWW+BRR+BWR)
\[\begin{gathered}
  { = ^3}{C_1}{(^2}{C_2}{ + ^4}{C_2}{ + ^2}{C_1}{ \times ^4}{C_1}) \\
   = 3 \times (1 + 6 + 8) \\
   = 45 \\
\end{gathered} \]
For step 2 we consider that 2 black balls are always included.
Step2:
Case 2: If we get 2 black balls then the third one is either red or white.
           = (BBW+BBR)
\[\begin{gathered}
  { = ^3}{C_2}{(^2}{C_1}{ + ^4}{C_2}) \\
   = 3 \times (2 + 4) \\
   = 18 \\
\end{gathered} \]
In the third case we consider all the balls to be black.
Step3:
Case 3: If all the balls are black.
\[{ = ^3}{C_3}\] = 1
Hence the number of ways is \[45 + 18 + 1 = 64\]


Note: In drawing a ball type problem, it becomes difficult for students as they mostly try to do it in a single step. The key is to be patient and try to form different cases and then solve it. Try to break it into small steps and then finally add them. Here at least means there is restriction for minimum but no restriction for maximum.