A box contains 2 white, 3 black and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least 1 black ball is to be included in the draw?
A. 32
B. 48
C. 64
D. 96
E. None of these
Answer: Option C
Solution(By Examveda Team)
We may have (1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).Required number of ways
$$\eqalign{ & \left( {{}^3{C_1} \times {}^6{C_2}} \right) + \left( {{}^3{C_2} \times {}^6{C_1}} \right) + \left( {{}^3{C_3}} \right) \cr & = \left\{ {3 \times \frac{{6 \times 5}}{{2 \times 1}}} \right\} + \left( {\frac{{3 \times 2}}{{2 \times 1}} \times 6} \right) + 1 \cr & = \left( {45 + 18 + 1} \right) \cr & = 64 \cr} $$
Related Questions on Permutation and Combination
A. 3! 4! 8! 4!
B. 3! 8!
C. 4! 4!
D. 8! 4! 4!
A. 7560,60,1680
B. 7890,120,650
C. 7650,200,4444
D. None of these
A. 8 × 9!
B. 8 × 8!
C. 7 × 9!
D. 9 × 8!
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