Question

In a box, there are $10$ balls, $4$ are red, $3$ black, $2$ white and $1$ yellow. In how many ways can a child select $4$ balls out of these $10$ balls? (Assume that the balls of the same colour are identical).

• A. $10$
• B. $15$
• C. $20$
• D. $25$

Answer 1

Answer: (C)

$20$

Let $x_1,x_2,x_3$ and $x_4$ be four balls selected of each color,
We just have to select four balls,
so, $x_1+x_2+x_3+x_4=4$$Max\left(x_1\right)=4$, $Max\left(x_2\right)-3$, $Max\left(x_3\right)=2$ and $Max\left(x_4\right)=1$
and
$min$ of any color is $0$$=(1+x+x^2+x^3+x^4)$ be the number of ways in which $0,1,2,3,4$ are selected from the red balls
$=(1+x+x^2+x^3)$ be the number of ways in which $0,1,2,3$ are selected from the black balls
$=(1+x+x^2)$ be the number of ways in which $0,1,2$ are selected from the white balls
$=(1+x)$ be the number of ways in which $0,1$ are selected from the yellow balls
So for selecting n balls:$=(1+x+x^2+x^3+x^4).(1+x+x^2+x^3).(1+x+x^2).(1+x)$
We want only the coeffiecient of $x^4$ as we have to select only $4$ balls
,$=(1+x+x^2+x^3+x^4).(1+x+x^2+x^3).(1+2x+2x^2+x^3)$
$=(1+x+x^2+x^3+x^4)(1+2x+2x^2+x^3+x+2x^2+2x^3+x^4+x^2+2x^3+2x^4+x^5+x^3+2x^4+2x^5+x^6)$
$=(1+x+x^2+x^3+x^4)(1+3x+5x^2+6x^3+5x^4+3x^5+x^6)$
$=5+6+5+3+1$ ways$=20$ ways
So, there are $20$ ways in which you can select $4$ balls.