Question

There are 4 routes between Delhi and Patna. In how many different ways can a man go from Delhi to Patna and return, if for returning
(i) Any of the routes is taken
(ii) The same route is taken
(iii) The same route is not taken?

Answer 1

Hint: Here we will apply the concept of multiplication in order to find out the possibilities. Also, with the help of the diagram the selection of routes will get easier. We will solve this question with and without repetition as per the parts have been given to us.

Complete step-by-step answer:
According to the question we are given 4 routes between Delhi and Patna. The diagram representing the four routes is given below.

A man can go by any of the routes he chooses. But the restriction is when the man returns from Patna to Delhi.
First we will consider the (i) case that says that the man can return from Patna to Delhi by any of the four routes denoted by ${{R}_{1}},{{R}_{2}},{{R}_{3}}$ and ${{R}_{4}}$.
So, first the man chooses any of the four routes from Delhi to Patna. This can be done in four ways only. Now we come to the limitation that he can return back from any one of the 4 routes. Let us suppose that he first went from the route ${{R}_{1}}$ and then he returns back from ${{R}_{1}}$ only. This results in only $1\times 1$ way. But if we focus on its possibilities then we have that he can also return by ${{R}_{2}}$ which can also be counted in the possibility. Continuing in this manner we see that going by route ${{R}_{1}}$ and then returning back gives $1\times 4$ ways. Similarly we have the same cases for ${{R}_{2}},{{R}_{3}}$ and ${{R}_{4}}$. Therefore, the total number of the cases will be added here. Thus, we get the possibilities as $\left( 1\times 4 \right)+\left( 1\times 4 \right)+\left( 1\times 4 \right)+\left( 1\times 4 \right)=16$ ways.
Now we will consider the second condition in which the limitation is given as he chooses to come back by the same route that he chose to go from Delhi to Patna. For example if he chooses to go from ${{R}_{1}}$ he comes back from ${{R}_{1}}$ only this can be done $1\times 1$ way. Similarly others are done as $1\times 1$ way. Therefore, the total number of possibilities here are $\left( 1\times 1 \right)+\left( 1\times 1 \right)+\left( 1\times 1 \right)+\left( 1\times 1 \right)=4$ ways.
Now, we will consider the third case in which he decides to come by the remaining routes which he had not chosen while going from Delhi to Patna. For example if he chooses the route ${{R}_{1}}$ then according to the case he will not return from ${{R}_{1}}$ but he will return from ${{R}_{2}},{{R}_{3}}$ and ${{R}_{4}}$. This can be done into $1\times 3$ ways. Similarly, we get other cases here as $1\times 3$ ways only. Therefore, the total number of possibilities is $\left( 1\times 3 \right)+\left( 1\times 3 \right)+\left( 1\times 3 \right)+\left( 1\times 3 \right)=12$ ways.

Note: We can also use the shortcut method for it. In case (i) this question can also be solved directly as going from any of the routes and coming back by any of the routes therefore, 4 routes multiplied by 4 ways of coming back .This results into $4\times 4=16$ ways. In case (ii) the ways are directly as $4\times 1=4$ ways. Similarly we can solve the third one also. While solving these questions one gets confused whether to solve it in addition or multiplication. Actually we add when one process stops but if the process is in continuation as we can see in this question then we do multiplication until the work is completed.