Q49 / 125 · BITSAT 2013 · Mathematics

A box contains two white balls, three black balls 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?

  • Correct answer
    A
    64
  • B
    129
  • C
    84
  • D
    None of these

Solution

Atleast one black ball can be drawn in the following ways (1) one black and two other colour balls $$ ={ }^{3} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{2}=3 \times 15=45 $$ (ii) two black and one other colour balls $$ ={ }^{3} \mathrm{C}_{2} \times{ }^{6} \mathrm{C}_{1}=3 \times 6=18 $$ (iii) All the three are black $={ }^{3} \mathrm{C}_{3} \times{ }^{6} \mathrm{C}_{0}=1$ $\therefore$ Req. no. of ways $=45+18+1=64$

Hence, the correct answer is option: 'A'.

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