Question

Six different balls are put in three different boxes, no box being empty. The probability of putting balls in the boxes in equal number is

• A. $\frac{3}{10}$
• B. $\frac{1}{6}$
• C. $\frac{1}{5}$
• D. None of these

Answer 1

The correct option is B $\frac{1}{6}$

Total number of ways to distribute the balls so that no box is empty are $\left[\right(1,1,4),(2,1,3),(2,2,2\left)\right]$
$=\frac{3!}{2!}\left[{}^6{C}_1{.}^5{C}_1{.}^4{C}_4\right]+3!\left[{}^6{C}_2{.}^4{C}_1{.}^3{C}_3\right]+\left[{}^6{C}_2{.}^4{C}_2{.}^2{C}_2\right]=90+6.60+90=540$
$\therefore$ Required probability $=\frac{90}{540}=\frac{1}{6}$