Question

There are 6 roads between A and B and 4 roads between B and C.
(i)In how many ways one can drive from A to C by way of B?
(ii)In how many ways can one drive from A to C and back to A, passing through B on both the trips?
(iii)In how many ways can one drive the circular trip described in (ii) without using the same road more than once.

Answer 1

Hint: We will use the fundamental principle of counting to calculate the ways on which one can drive and then we will consider the given conditions to solve the questions. We will calculate the required ways one by one for each part like in part (i) we will find the ways which one can go through from A to C but with the condition that one must cross through B while going from A to C. If there are 6 ways between A and B, so they can choose any one out of those 6 and ways from B to C are 4, so again, they can choose any one out of 4. The total will be $6 \times 4$ using the fundamental principle of counting.

Complete step-by-step answer:
We are given that roads between A and B are 6 and roads between B and C are 4.
 For option(i), we have to find the number of ways in which one can drive from A to C given the condition that they must travel through B.
So, for A to B one can travel through 6 roads and from B to C, one can have 4 roads to travel through.
We can define the fundamental principle of counting as: a rule which is used to calculate the total number of possible outcomes of any situation. If there are m ways of doing a given situation and n ways to do a situation simultaneously after that, then it states that there are total mn ways to do both the actions.
Therefore, the total number of ways using the fundamental principle of counting are: $6 \times 4 = 24$
Hence, one can have 24 ways to drive from A to C by way of B.
For option(ii), we are given that one travels from A to C and then back to A passing through B on both the trips i.e., their trip can be shown as $A \to B \to C \to B \to A$
So from A to B, there are 6 roads, from B to C there are 4 roads and from C to B there are again 4 roads and from B to A, there are 6 possible roads. Using the fundamental principle of counting, we get the total number of ways as:$6 \times 4 \times 4 \times 6 = 576$
Hence, one can have 576 ways from A to C and then back to A passing through B on both the trips.
For option(iii), we have to find the same ways as we did in part (ii) but without repeating the roads.
Here, for the circular trip $A \to B \to C \to B \to A$, we can have 6 roads from A to B, 4 roads from B to C but 3 roads when coming back from C to B because one can use a road once only and similarly 5 roads from B to A.
Using fundamental principle of counting, we get the total number of ways as : $6 \times 4 \times 3 \times 5 = 360$
Therefore, there are 360 ways for the circular trip $A \to B \to C \to B \to A$.

Note: Such questions are confusing so it’d be advisable to draw the sketch route in rough space while using the fundamental principle of counting. You may get confused in the use between the fundamental principle of counting and the permutations method.
Fundamental principle of counting: It is a way to calculate the possible number of outcomes such that if there are p ways to do one thing and q ways to do the other, then total ways to do both the things are pq.
Permutations: In mathematics, an arrangement of all elements of a set with respect to the order of the arrangement and it is used for lists and order of elements of a set is taken into account.