A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one

Q: Wait, why do we multiply by $3^2$ for the incorrect questions? Shouldn't it be $1$ way to get it wrong?

That is a common trap! Remember, each question has $4$ options. Only $1$ is correct, which leaves $3$ options that are incorrect. If you choose an incorrect option, you have $3$ distinct choices for that specific question. Since there are $2$ questions that must be answered incorrectly, you have $3 \times 3 = 3^2$ ways to fill those slots.

Q: Can't we just use probability here? Like $P = ^6C_4 \times (1/4)^4 \times (3/4)^2$?

You are thinking of the Binomial Distribution, which calculates the probability of an event. However, this question asks for the 'number of ways,' which is a counting problem. We are not looking for the likelihood; we are looking for the total count of unique combinations of answers. So, we ignore the probability fractions and focus only on the number of choices.

Q: What if the question said 'at least' four correct answers? Would the logic change?

Absolutely! 'Exactly' is a fixed constraint. 'At least' would mean you have to calculate the cases for exactly $4$, exactly $5$, and exactly $6$ correct answers separately, and then add them together. It would be a much longer problem, but the fundamental logic of using combinations and powers would remain the same.

Question

A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is

Sigma Percentile · Accepted Answer

### The Architecture of Choice: Mastering Combinatorics in Exams

Imagine you are sitting in the exam hall. The clock is ticking, the pressure is mounting, and you are staring at a test paper with $6$ multiple-choice questions.

Each question has $4$ options, but only one is the golden ticket—the correct answer. The other $3$ are decoys.

The problem asks us to find the number of ways to answer this test such that exactly $4$ questions are correct. This is not just a math problem; it is a lesson in structured decision-making.

### Phase 1: The Selection of Destiny

Before we worry about the options, we must address the structure. We have $6$ questions, and we need to decide which $4$ of them will be the 'correct' ones.

The order in which we select these questions does not matter; we are simply choosing a set of $4$ positions out of $6$. This is the classic domain of combinations, using the formula $^nC_r$ where $n=6$ and $r=4$.

Using the symmetry property $^nC_r = ^nC_{n-r}$, we know that $^6C_4$ is identical to $^6C_2$. The calculation is as follows:

$$^6C_4 = ^6C_2 = \frac{6 	imes 5}{2 	imes 1} = 15$$

So, there are $15$ distinct ways to choose which questions will be correct.

### Phase 2: The Precision of Correctness

Now that we have selected our $4$ questions, we must answer them. Since each question has only $1$ correct option, there is only $1$ way to answer each of these $4$ questions correctly.

If we have $4$ questions, the number of ways to answer them is:

$$1 	imes 1 	imes 1 	imes 1 = 1^4 = 1$$

It seems trivial, but in combinatorics, acknowledging the 'one way' is vital for maintaining the integrity of our logic.

### Phase 3: The Trap of the Incorrect

This is where most students stumble. We have $2$ questions remaining ($6 - 4 = 2$). The problem demands *exactly* $4$ correct answers, which implies that these remaining $2$ questions *must* be answered incorrectly.

For each of these $2$ questions, there are $4$ options total. Since $1$ is correct, there are $4 - 1 = 3$ incorrect options.

By the Fundamental Principle of Counting, the number of ways to answer these $2$ questions incorrectly is:

$$3 	imes 3 = 3^2 = 9$$

### The Synthesis: Bringing It All Together

We have three independent events that must occur simultaneously: selecting the questions, answering the correct ones, and answering the incorrect ones. To find the total number of ways, we multiply these independent counts together:

$$15 	imes 1 	imes 9 = 135$$

The final calculation is straightforward: $15 	imes 9 = 135$. There are exactly **$135$** ways to answer the test to satisfy the condition.

The beauty of this problem lies in the realization that 'exactly' constraints are not just about what you include, but also about what you must strictly exclude. Keep this in mind, and you will never fall into the trap of overcounting again.

Animated Solution for Mathematics - Permutations and Combinations: A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is

Enter Numerical Value:

Visualized Solution

Visualizing the Test Structure

Selecting the Correct Questions

Applying the Combination Formula

Calculating $^{6} C_{4}$

Answering the Correct Questions

Ways to Answer Correctly

The Trap of the Remaining Questions

Ways to Answer Incorrectly

Combining All Independent Events

Final Calculation

The Sigma Insight: Combinations and Selection

The Architecture of Choice

Phase 1

Phase 2

Phase 3

The Synthesis

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