- Home
- Standard 11
- Mathematics
6.Permutation and Combination
medium
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
A
$64$
B
$45$
C
$46$
D
None of these
(IIT-1986)
Solution
(a) $A$ 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.
Standard 11
Mathematics
