
The number of ways in which n different objects can be put into 2 different boxes is
Answer
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Hint: We will take objects one by one and we can see for each object, we have 2 options box 1 and box 2. We will find the total ways by applying the same procedure one by one on all other objects.
Complete step by step solution: We have to find the number of ways in which n different objects can be put into 2 different boxes.
Let there are 2 different boxes A and B and let the n different objects be $1,2,3,4,...,n$
Let’s choose the first object,
We can see the object has 2 option where to be put in, either Box A or Box B.
So, the number of options of putting 1st object: 2
Now, for the second object, it doesn’t matter in which box first object was placed.
So the events of putting successive objects will be independent.
It is also necessary that all objects are to be placed in box 1 or box 2, so it doesn’t matter which object is placed first and which later, the only thing that matters is the final outcome, i.e. after placing n objects
We can see for each object chosen, there are 2 options for each object.
Thus, for n objects, there will be $2 \times 2 \times 2 \times 2...ntimes$ number of ways.
So, total ways: $2 \times 2 \times 2 \times 2...ntimes = {2^n}$
Hence, there are ${2^n}$ number of ways in which n different objects can be put into 2 different boxes.
Note: It must be seen carefully in the question of whether the objects and boxes are both different and identical. A different approach to the question can be assuming there are $r$ objects in box 1 and $\left( {n - r} \right)$objects in box 2 at the end of the experiment.
So, the number of ways of choosing $r$ objects out of n objects is ${}^n{C_r}$
The $\left( {n - r} \right)$objects in box 2 will be chosen automatically.
Now, objects in box 1 can vary from 0 to n, thus $r$can vary from 0 to n.
So, total ways: $\sum\limits_{r = 0}^n {{}^n{C_r}} = {2^n}$
Hence, the same answer.
Complete step by step solution: We have to find the number of ways in which n different objects can be put into 2 different boxes.
Let there are 2 different boxes A and B and let the n different objects be $1,2,3,4,...,n$
Let’s choose the first object,
We can see the object has 2 option where to be put in, either Box A or Box B.
So, the number of options of putting 1st object: 2
Now, for the second object, it doesn’t matter in which box first object was placed.
So the events of putting successive objects will be independent.
It is also necessary that all objects are to be placed in box 1 or box 2, so it doesn’t matter which object is placed first and which later, the only thing that matters is the final outcome, i.e. after placing n objects
We can see for each object chosen, there are 2 options for each object.
Thus, for n objects, there will be $2 \times 2 \times 2 \times 2...ntimes$ number of ways.
So, total ways: $2 \times 2 \times 2 \times 2...ntimes = {2^n}$
Hence, there are ${2^n}$ number of ways in which n different objects can be put into 2 different boxes.
Note: It must be seen carefully in the question of whether the objects and boxes are both different and identical. A different approach to the question can be assuming there are $r$ objects in box 1 and $\left( {n - r} \right)$objects in box 2 at the end of the experiment.
So, the number of ways of choosing $r$ objects out of n objects is ${}^n{C_r}$
The $\left( {n - r} \right)$objects in box 2 will be chosen automatically.
Now, objects in box 1 can vary from 0 to n, thus $r$can vary from 0 to n.
So, total ways: $\sum\limits_{r = 0}^n {{}^n{C_r}} = {2^n}$
Hence, the same answer.
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