Question

A question paper consists of 6 questions, each having an alternative. The number of ways an examinee can answer one or more questions is
(a) 243
(b) 242
(c) 729
(d) 728

Answer 1

Each question has 3 possibilities, that is, either answer the first alternative or the second alternative or leave the question unattempted. These are applicable for 6 questions. Therefore, the total combination will be $3\times 3\times 3\times 3\times 3\times 3={{3}^{6}}$ . According to the given condition, we should not consider the case when the question is unattempted. We can find the number of ways an examinee can answer one or more questions by subtracting the total number of ways the questions are unattempted from the total number of ways. First question has a chance of being unattempted 1 time, the second question will have a chance of being unattempted 1 time and so on. Therefore, we can write the number of ways this can be done as $1\times 1\times 1\times 1\times 1\times 1=1$ . Now, we have to substitute the values and simplify.

Complete step by step answer:
We have to find the number of ways a student can answer one or more questions from6 questions, each having an alternative. We can see that for each question, there are 3 possibilities, that is, either answer the first alternative or the second alternative or leave the question unattempted. These are applicable for 6 questions. Therefore, we can find the total combination as
$3\times 3\times 3\times 3\times 3\times 3={{3}^{6}}=729$
We are given that the student must answer one or more questions. Therefore, we should not consider the case when the question is unattempted.
Therefore, we can find the number of ways an examinee can answer one or more questions by subtracting the total number of ways the questions are unattempted from the total number of ways.
$\Rightarrow$ Required number of ways = Total number of ways - Total number of ways the questions are unattempted (Because given that at least need to answer one question)...(i)
Let us find the total number of ways the questions are unattempted. We have seen that each question has 3 possibilities, that is, either answer the first alternative or the second alternative or leave the question unattempted. So, the first question has a chance of being unattempted 1 time, the second question will have a chance of being unattempted 1 time and so on. Therefore, we can write the number of ways this can be done as
$1\times 1\times 1\times 1\times 1\times 1=1$
Now, let us substitute the values in (i).
$\Rightarrow \text{Required number of ways}=729-1=728$

So, the correct answer is “Option d”.

Note: Students have a chance of making a mistake by considering the possibility of each question as first alternative or the second alternative and thereby writing the total combination as ${{2}^{6}}$ . Students must never forget to subtract the total number of ways the questions are unattempted from the total number of ways.