Question

A box contains 2 white balls, 3 black balls and 4 red balls. 3 balls are drawn from the box such that there is at least 1 black ball in the drawn balls. What is the number of ways of drawing such 3 balls from the box?
(a) 6, if the balls of the same colour are alike.
(b) 8, if the balls of the same colour are alike.
(c) 64, if balls of the same colour are different.
(d) 80, if balls of the same colour are different.

Answer 1

Hint: This question is based on the concept of number of non-negative integral solutions of a linear multi variable equation. Form such an equation and then apply the formula.
If suppose there are n variables x1, x2, x3, . . . , xn-1, xn which takes only integer values such that the following equation satisfy
x1 + x2 + x3 + . . . + xn-1 + xn = m
and xi ≥ 0 for all i such that 1 ≤ I ≤ n and m, n $\in {{\mathbb{Z}}^{+}}$(non-negative integers)
Then the number of solutions of the above equation is given by
$N=\dfrac{\left( m+n-1 \right)!}{m!\left( n-1 \right)!}$
where N is the total number of solutions.

Complete step-by-step answer:

If suppose there are n variables x1, x2, x3, . . . , xn-1, xn which takes only integer values such that the following equation satisfy
x1 + x2 + x3 + . . . + xn-1 + xn = m
and xi ≥ 0 for all i such that 1 ≤ I ≤ n and m, n $\in {{\mathbb{Z}}^{+}}$(non-negative integers)
Then the number of solutions of the above equation is given by
$N=\dfrac{\left( m+n-1 \right)!}{m!\left( n-1 \right)!}$
where N is the total number of solutions.
Case 1: When balls of the same colour are alike.
Let x, y and z denote the number of white, black and red balls respectively drawn from the box. The question says that 3 balls are drawn. So we get the equation as:
$x+y+z=3\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(i)}$
Now it is given that at least 1 black ball is to be included in the draw. That means
$y\ge 1$
Also there is a limited number of each kind of ball. So from information in question we get
$\begin{aligned} & 0\le x\le 2 \\\ & 1\le y\le 3 \\\ & 0\le z\le 4 \\\ \end{aligned}$
We see that the constraint is not of the required form.
Let $y=t+1$ . Then $1\le y\le 3\Rightarrow 0\le t\le 2$
Replacing y with t in equation (i) we get
$\begin{aligned} & x+t+1+z=3 \\\ & \Rightarrow x+t+z=2\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(ii)} \\\ \end{aligned}$
and
$\begin{aligned} & 0\le x\le 2 \\\ & 0\le t\le 2 \\\ & 0\le z\le 4 \\\ \end{aligned}$
There is no need to consider the RHS constraints in x, t and z because equation (ii) already imposed that condition.
So our equation becomes
$\begin{aligned} & x+t+z=2 \\\ & \text{with} \\\ & x\ge 0 \\\ & t\ge 0 \\\ & z\ge 0 \\\ \end{aligned}$
Now applying the formula of number of integral solutions we get
n = 3 , m = 2
$\begin{aligned} & \text{So}\,,\,N=\dfrac{(2+3-1)!}{2!(3-1)!} \\\ & \Rightarrow N=\dfrac{4!}{2!2!}=\dfrac{24}{2\times 2}=6 \\\ \end{aligned}$
Hence there are 6 ways to draw 3 balls from the box.
So, option (a) is correct.
Case 2: When balls of the same colour are different.
Let the balls be named as ${{W}_{1}}\,,\,{{W}_{2}}\ ,\ {{B}_{1}}\ ,\ {{B}_{2}}\ ,\ {{B}_{3}}\ ,\ {{R}_{1}}\ ,\ {{R}_{2}}\ ,\ {{R}_{3}}\ ,\ {{R}_{4}}$ where W, B, R denote white, black and red balls.
First let us count the total number of ways of drawing 3 balls with no restriction. Let M denote that number of ways. Then it is equivalent to choose three balls among the 9 given.
$M=\dfrac{9!}{6!3!}=\dfrac{9\times 8\times 7\times 6!}{6!\times 6}=84$
Now we will find the number of ways of drawing 3 balls such that no black ball is drawn. Let the N denote that number. Then it is equivalent to choose three balls among the 6 balls (excluding black balls).
$N=\dfrac{6!}{3!3!}=\dfrac{6\times 5\times 4\times 3!}{3!\times 6}=20$
To get the number of ways of drawing the balls such that at least one black ball is drawn we will subtract the number of draws excluding black balls(N) from the total number of draws(M). Let K denote that number. Then
$K=M-N=84-20=64$
So, there are 64 ways of drawing the 3 balls from the box.
Hence option (c) is correct.
Hence, option (a) and (c) both are correct.

Note: One may confuse how to deal with RHS of constraint of x, t and z and without removing those RHS constraint, systems of that form cannot be achieved. It is also important to consider both the cases, when the balls are alike and when they are different. Both of these give different answers and hence, give us two correct options.