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# A question paper has $5$ questions. Each question has an alternative. The number of ways in which a student can attempt at least one question is?A.${2^5} - 1$ B. ${3^5} - 1$ C.${3^4} - 1$ D.None of these

Last updated date: 12th Sep 2024
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Hint: 1. In such types of questions it is important for the students to understand the permutation and combination terms given.
2. In this question students need to classify the data as the events mentioned and total outcomes.
3. The tricky part in this is to establish the relationship between the events and the outcomes.
To answer this question, we must have an idea about permutation and combination. Permutation is an act of arranging the objects or numbers in order. Combinations are the way of selecting the objects or numbers from a group of objects or collections, in such a way that the order of the objects does not matter.

let us understand what the question provides and what it demands.
The question says,A question paper has $5$ questions. Each question has an alternative.
It demands us to find,The number of ways in which a student can attempt at least one question is?
From the formula mentioned and understanding the given data we can say,
The student can attempt each question using any of two options or alternatives or can choose not to answer that question total there are $5$ questions.
hence the number of ways in which a student can attempt at least one question is $\left( {3 \times 3 \times 3 \times 3 \times 3} \right) - 1 = {3^5} - 1$.
Here, we have taken 3 because the student has 3 choices,
To attempt the first question of the alternative given.
To attempt a second question of the alternative given.
To skip both the alternative questions.
We need to subtract one $i.e.,{\text{ }}1$case where the student opted to not attempt to answer every question.

So, the correct answer is “Option B”.

Note: In this type of problem-related to permutation and combination, we should know about the factorial rotation$,{\text{ }}i.e.,$factorial $n$ is denoted as $n!$ and is defined as $n! = n\left( {n - 1} \right)\left( {n - 2} \right).....3.2.1$.
Number of permutations is defined as $n\Pr = n\left( {n - 1} \right)\left( {n - 2} \right)......\left( {n - r + 1} \right) = \dfrac{{n!}}{{\left( {n - r} \right)!}}$.
Number of combinations is defined as $nCr = \dfrac{{n!}}{{r!}}\left( {n - r} \right)! = n\left( {n - 1} \right)\left( {n - 2} \right)......to{\text{ }}\dfrac{{r{\text{ }}factors}}{{r!}}$.
Students must understand what is asked in such a question and should be able to classify the data correctly.
Students might mess up while putting the values in the actual equation.