Question

There are 6 multiple choice questions in an examination. Find the total number of ways of answering these questions if the first three questions have 5 choices each and the next three have 4 choices each?

Answer 1

Hint: We will use the concept of multiplication as here the question is about finding the possibilities of a continuation process that is answering the question paper. With the help of this concept we will find the ways for the question.

Complete step-by-step answer:
According to the question we have a multiple choice question in which we are given 6 questions along with their options. In this question paper the number of choices given has been divided into two sets. This means that the first three questions have 5 choices and the next three questions carry 4 choices. So, now we will first consider those questions which have 5 choices in them. So, any one question out of these has 5 choices therefore can be answered in 5 ways. Thus, we have $1\times 5$ ways of answering one question. Thus, this is similar to the two questions also. So the number of possibilities here is given by $\left( 1\times 5 \right)\times \left( 1\times 5 \right)\times \left( 1\times 5 \right)$. Multiplication has been done here because the answering process is not yet stopped. This is due to the fact that he has 6 multiple choice questions in total.
Now we will consider the questions which carry 4 choices with them. Now we will consider any one question here. Suppose we have chosen one question out of these types and we need to answer it. So, this can be done in $1\times 4$ ways. Similarly others questions are also done like this. Therefore, the total number of ways are $\left( 1\times 4 \right)\times \left( 1\times 4 \right)\times \left( 1\times 4 \right)$. And since all the answering process is in the continuing process therefore the total number of possibilities is $\left( 1\times 5 \right)\times \left( 1\times 5 \right)\times \left( 1\times 5 \right)\times \left( 1\times 4 \right)\times \left( 1\times 4 \right)\times \left( 1\times 4 \right)=5\times 5\times 5\times 4\times 4\times 4$. Or we can say that the total possibilities here are 8000.

Note: One should not get confused while using plus and minus signs between the numbers here. The explanation for this is that if the work is in continuation then the numbers are followed by a multiplication sign among them. Otherwise, if the work is completed and then started with a new scale then addition will take place. As in this question answering the question paper is a continuation process. Therefore, multiplication is applied between the numbers.