A) 64
B) 45
C) 46
D) None of these
Correct Answer: A
Solution :
A selection of 3 balls so as to include at least one black ball, can be made in the following 3 mutually exclusive ways (i) 1 black ball and 2 others = \[^{3}{{C}_{1}}\times {{\,}^{6}}{{C}_{2}}=3\times 15=45\] (ii) 2 black balls and one other =\[^{3}{{C}_{2}}\times {{\,}^{6}}{{C}_{1}}=3\times 6=18\] (iii) 3 black balls and no other = \[^{3}{{C}_{3}}=1\] \[\therefore \] Total numbers of ways = 45 + 18 + 1 = 64.
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