Question

If $5$ different things are placed at random in $3$ different boxes then the number of ways of placing them such that no box remains empty is

• A. $150$
• B. $120$
• C. $155$
• D. $134$

Answer 1

The correct option is A $150$

· No. of ways to choose the box which gets 3 balls $=3$ (since, 3 boxes are there).

· No. of ways to select 3 balls out of 5 is ${=}^5{C}_3$.

· No. of ways to distribute rest of the balls $=2.$

· Total no. of ways $=60$

· No. of ways to select 1 balls out of 5 is ${=}^5{C}_1.$

· No. of ways to choose the box which gets 1 balls $=3$ (since, 3 boxes are there).

· No. of ways to select 2 balls out of remaining 4 to go in second box is = ${}^4{C}_2$.

· Total no. of ways $=90$

Total: $60+90=150.$