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**Hint:**: In this question, it has been asked that there are three multiple-choice questions and we have four choices in each question. So, to calculate the no. of ways by which a student can get all answers correct, first of all, find the total no. of ways in which a student can give answers that can be obtained by multiplying no. of options for each question $(4)$ for no. of times that is equal to no. of questions $(3)$. Then calculate no. of ways there can be to get the correct answer. Finally, subtract no. of ways of getting the correct answer from the total no. of ways of getting the answer to get no. of ways by which a student can fail to get all answers correct.

**Complete step by step solution:**

First, we have to calculate the total no. of ways to solve a correct question –

Total no. of ways in which answers can be given $ = 4 \times 4 \times 4$

$ \Rightarrow 64$ ….[as each question has four choices]

Then we have to calculate the total no. of ways to have a correct answer.

So, the total no. of ways to have a correct answer will be:

$ = 1 \times 1 \times 1$$ = 1$…...[among three multiple-choice question, each has $1$ correct answer.

So, the no. of ways that a student fails to correct an answer will be

Total no. of way student fails to correct = total no. of ways – no. of ways student corrects it.

$ = 64 - 1$

$ = 63$

**$\therefore$ Number of ways in which a student can fail to get all answer correct is 63. Hence the correct option is (D).**

**Note:**

Read the question very carefully to understand what is asked as the ultimate answer because there is a high chance of getting confused between whether it is asked for no. of ways for getting all correct or incorrect answer & a little ignorance here can lead to an incorrect answer even after knowing the whole procedure to do the sum. Do the calculations attentively. This method is the easiest way to solve this type of question. This procedure can be done in so limited time too so it’s helpful for solving just in mind within seconds too.

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