JEE Main 2022 (24 June Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Permutations and Combinations: In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, -2 marks for each wrong answer and 0 mark if the question is not attempted. Then, the number of ways a student appearing in the examination gets 5 marks is_.

Q: Wait, why do we reject the negative root or non-integer solutions?

Great question! In this problem, $c$ (correct) and $w$ (wrong) represent the count of questions. You cannot have a negative number of questions or a fraction of a question. Therefore, we are strictly looking for non-negative integer solutions to our system of equations. If a value for $c$ or $w$ results in a negative number or a fraction, it is physically impossible in the context of the exam.

Q: What if $u$ (unattempted) was not zero? Does the formula change?

The logic remains robust! If $u$ were greater than zero, you would simply include it in your combinatorial selection. You would choose $c$ questions to be correct, $w$ questions to be wrong, and the remaining $u$ questions would be unattempted. The total ways would then be $\binom{5}{c, w, u} \times (1)^c \times (2)^w \times (1)^u$. In this specific case, $u=0$, so it simplifies nicely.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the exam hall. You have 5 questions. Let

c

w

, and

u

be the number of correct, wrong, and unattempted questions. You must satisfy two simultaneous conditions: the total number of questions

c + w + u = 5

and the total marks

3 c - 2 w + 0 u = 5

. The key is to realize that

c, w, u

must be non-negative integers.

Hint 2

Hint 2 (The Arsenal): Once you find the valid integer solution for

(c, w, u)

, you are dealing with a selection problem. You need to choose which specific questions are correct and which are wrong. Use the combination formula

(k n)

to select the questions, and then account for the number of ways to answer each type of question (1 way for correct, 2 ways for wrong).

Hint 3

Hint 3 (The Execution): Start by solving

3 c - 2 w = 5

for integer values of

c

and

w

where

0 \leq c + w \leq 5

. Once you have the unique set

(c, w, u)

, calculate the total ways using the formula:

(c 5) \times (1)^{c} \times (2)^{w}

Visualized Solution

Define the Problem Variables

Total questions: $n = 5$
Choices per question: $3$
Marks: Correct ( $+ 3$ ), Wrong ( $- 2$ ), Unattempted ( $0$ )
Target Total Marks: $5$

Set up the Mathematical Equations

Let $c, w, u$ be the number of correct, wrong, and unattempted questions.
Constraint 1 (Total Questions): $c + w + u = 5$
Constraint 2 (Total Marks): $3 c - 2 w + 0 u = 5$

Analyze the Marks Equation

Focus on: $3 c - 2 w = 5$
Since $c, w \geq 0$ , test integer values for $c$ .

Find the Valid Combination

If $c = 3$ : $3 (3) - 2 w = 5$
$9 - 2 w = 5 \Rightarrow 2 w = 4 \Rightarrow w = 2$
Total questions: $c + w = 3 + 2 = 5 \Rightarrow u = 0$
Valid Case: $(3, 2, 0)$

Select the Correct Questions

We need exactly $3$ correct questions out of $5$ .
Number of ways to choose: $(3 5)$
$(3 5) = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$ ways

Ways to Answer Correct Questions

For each correct question, there is only $1$ correct choice out of $3$ .
Ways to answer $3$ correct questions: $1 \times 1 \times 1 = 1^{3} = 1$ way

Ways to Answer Wrong Questions

For the remaining $2$ questions, they must be wrong.
Out of $3$ choices, $1$ is correct, so $3 - 1 = 2$ choices are wrong.
Ways to answer $2$ wrong questions: $2 \times 2 = 2^{2} = 4$ ways

Calculate Total Number of Ways

Total ways = (Choose 3 questions) $\times$ (Ways for correct) $\times$ (Ways for wrong)
Total ways = $10 \times 1^{3} \times 2^{2}$
Total ways = $10 \times 1 \times 4 = 40$

00:00 / 00:00

The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

To solve this problem, we define our variables based on the scoring system. Let

c

be the number of correct answers,

w

be the number of wrong answers, and

u

be the number of unattempted questions.

We are governed by two fundamental constraints: 1. The total number of questions is

5

c + w + u = 5

2. The total marks must be

5

3 c - 2 w = 5

Solving the Integer Constraints

We must find non-negative integers

c

and

w

that satisfy the equation

3 c - 2 w = 5

. We test possible values for

c

c = 1

, then

3 (1) - 2 w = 5

, which leads to

2 w = - 2

(impossible).

c = 2

, then

3 (2) - 2 w = 5

, which leads to

2 w = 1

(impossible).

c = 3

, then

3 (3) - 2 w = 5

, which leads to

2 w = 4

, so

w = 2

. This is a valid solution.

Since

c = 3

and

w = 2

, the sum

c + w = 5

. Substituting this into our first constraint, we find

u = 0

Calculating the Combinations

Now that we have determined the counts (

3

correct,

2

wrong,

0

unattempted), we calculate the number of ways to arrange these outcomes. We must choose

3

questions out of

5

to be correct:

(3 5) = \frac{5 !}{3 ! 2 !} = 10

For the

3

correct questions, there is only

1

way to answer each correctly. Thus, we have

1^{3} = 1

way.

For the

2

wrong questions, assuming there are

3

total options per question (one correct, two incorrect), there are

2

incorrect options for each wrong answer. Thus, we have

2^{2} = 4

ways.

Final Calculation

To find the total number of ways to achieve a score of

5

, we multiply the combinations by the ways to answer the questions:

Total Ways = 10 \times 1 \times 4 = 40

The total number of ways to score exactly

5

marks is $40$ .

Common Traps & FAQs

Wait, why do we reject the negative root or non-integer solutions?

Great question! In this problem,

c

(correct) and

w

(wrong) represent the count of questions. You cannot have a negative number of questions or a fraction of a question. Therefore, we are strictly looking for non-negative integer solutions to our system of equations. If a value for

c

w

results in a negative number or a fraction, it is physically impossible in the context of the exam.

Why is it

2^{w}

for the wrong answers and not

3^{w}

Remember, each question has 3 choices. Only 1 choice is correct. If the student gets the question wrong, they must have chosen one of the *other* options. Since there are 3 total options and 1 is correct, there are

3 - 1 = 2

incorrect options. Thus, for each wrong question, there are 2 ways to answer it incorrectly.

What if

u

(unattempted) was not zero? Does the formula change?

The logic remains robust! If

u

were greater than zero, you would simply include it in your combinatorial selection. You would choose

c

questions to be correct,

w

questions to be wrong, and the remaining

u

questions would be unattempted. The total ways would then be

(c , w , u 5) \times (1)^{c} \times (2)^{w} \times (1)^{u}

. In this specific case,

u = 0

, so it simplifies nicely.

Enter Numerical Value:

Visualized Solution

Define the Problem Variables

Set up the Mathematical Equations

Analyze the Marks Equation

Find the Valid Combination

Select the Correct Questions

Ways to Answer Correct Questions

Ways to Answer Wrong Questions

Calculate Total Number of Ways

The Sigma Insight: Combinations and Selection

Analyzing the Setup

Solving the Integer Constraints

Calculating the Combinations

Final Calculation

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