Question

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 atleast one black ball is to be included in the draw?

Answer 1

Number of ways of selecting ‘r’ different objects from ‘n’ distinct objects is ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
n! is defined as n factorial which is equal to product of all positive integers less than or equal to n.

Complete step-by-step answer:
The box contains two white, three black and four red balls. Number of ways of selecting three balls out of the box such that at least one black ball is included:
So, we have to make different cases:
Case-1: When there is exactly one black ball and two non black(red or white) balls.
Number of ways to draw one black ball from three black balls is ${}^3{C_1}$
Number of ways of selecting two non black balls from total six non black balls: ${}^6{C_2}$
Total number of ways in case 1: ${}^3{C_1} \times {}^6{C_2}$ ---(1)
Case-2: When there is exactly two black balls and one non black ball (red or white)
Number of ways to draw two black balls from three black balls is ${}^3{C_2}$
Number of ways of selecting one non black balls from total six non black balls: ${}^6{C_1}$
Total number of ways in case 2: ${}^3{C_2} \times {}^6{C_1}$ ---(2)
Case-3: When there is all three black balls
Number of ways to draw three black balls from three black balls is ${}^3{C_3}$ -- (3)
From equation (1), (2) and (3); we get
Total number of ways: ${}^3{C_1} \times {}^6{C_2} + {}^3{C_2} \times {}^6{C_1} + {}^3{C_3}$ --(4)
${}^3{C_1} = \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} = \dfrac{6}{{1 \times 2}} = 3$
${}^6{C_2} = \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}} = \dfrac{{6 \times 5}}{{1 \times 2}} = 15$
${}^3{C_2} = \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} = \dfrac{6}{{1 \times 2}} = 3$
${}^6{C_1} = \dfrac{{6!}}{{1!\left( {6 - 1} \right)!}} = 6$
${}^3{C_3} = \dfrac{{3!}}{{3!\left( {3 - 3} \right)!}} = 1$
Putting all values in equation (4);
Total number of ways = $(3 \times 15) + (3 \times 6) + 1 = 45 + 18 + 1 = 64$

Note: We can also find number of ways of selecting at least one black balls = Total number of ways of selecting 3 balls – Number of ways of selecting 3 non- black balls
Number of ways = ${}^9{C_3} - {}^6{C_3} = \dfrac{{9!}}{{3!\left( {9 - 3} \right)!}} - \dfrac{{6!}}{{3!\left( {6 - 3} \right)!}} = 84 - 20 = 64$