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Question

$$5$$ balls are to be placed in $$3$$ boxes. Each box can hold all the $$5$$ balls. Number of ways in which the balls can be placed so that no box remains empty, if :


Solution

$$A.$$
Each box (say $$B_1,\, B_2,\, B_3$$) will have at least one ball.

Now the ways for placing other $$2$$ identical balls in $$3$$ different boxes are

$$\displaystyle \frac {(2+3-1)!}{2!(3-1)!}\, =\, 6\quad \left (\therefore \, \displaystyle \frac {(n+r-1)!}{n!(r-1)!} \right )$$
$$B.$$
Case $$1$$ : $$5$$ balls can be divided in $$3$$ groups having $$2$$ balls each in $$2$$ boxes and $$1$$ ball for in third box $$(2, 2, 1)$$

ways : $$\displaystyle \frac {5!}{(1!)(2!)^2\, \times 2!}\, =\, 15$$

Case $$2$$ : Division can also be $$3$$ in one box and $$1$$ each in remaining $$2$$ boxes $$(3, 1, 1)$$

ways : $$\displaystyle \frac {5!}{(3!)(1!)^2\, \times 2!}\, =\, 10$$


Hence total ways $$= 10 + 15 = 25$$
$$C.$$
Only $$2$$ arrangements are possible.

1. $$2$$ balls each in $$2$$ boxes & remaining ball in other box $$(2, 2, 1)$$

2. $$3$$ balls in $$1$$ box and 1 ball each in other boxes $$(3, 1, 1)$$

$$D.$$
Same as part $$A$$

Maths

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