Question

find the number of ways in which n distinct balls can be put into three boxes so that no two boxes remain empty​

Answer 1

Number of ways in which n distinct balls can be put into three boxes so that no two boxes remain empty​ = 3ⁿ - 3

Step-by-step explanation:

Let say three boxes A , B & C

Each ball can be placed in 3 Ways

number of ways in which n distinct balls can be put into three boxes

= 3ⁿ

now Take The case when two boxes remain Empty

2 boxes out of 3 can be empty  

in ³C₂  =  3 Ways    ( AB  , BC  , AC)

when these boxes are empty then n Balls can be put in 1 Way only

number of ways in which n distinct balls can be put into three boxes so that no two boxes remain empty​ = 3ⁿ - 3

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Answer 2

Given: $n$ distinct balls and three boxes.

To Find: Find the number of ways in which $n$ distinct balls can be put into three boxes so that no two boxes remain empty​.

Step-by-step explanation:

For every object we have three options, putting them in either of the three boxes.

Thus,

The objects can be put in $3^{n}$ ways.

Since,

The boxes which can be arranged among them self in $3!$ ways are identical.

Therefore,

The number of ways$=\frac{3^{n} }{3!}$

But this includes one case in which all the object are put in one box.

It also include the $(2^{n-1}-1)$ cases in which all the objects are put in two boxes only.

Hence,

Required number of ways is,

$=\frac{3^{n} }{3!}-1-(2^{n-1}-1)$

$=3^{n-1}-2^{n-1}$

So, The Answer is $(3^{n-1}-2^{n-1})$.