Question

An examination consists of $10$ multiple choice questions, where each question has $4$ options, only one of which is correct. In every question, a candidate earns $3$ marks for choosing the correct option, and $-1$ for choosing a wrong option.

Assume a candidate answers all questions by choosing exactly one option for each. Then the number of distinct combinations of answers which can earn the candidate a score from the set $\{15,16,17,18,19,20\}$ is

• A. $27$
• B. $120$
• C. $3240$
• D. $6480$

Answer 1

The correct option is C $3240$

Exam consists of $10$ multiple choice questions.Whre out of $4$ option one is correct and $3$ are wrong.

Candidate get $3$ marks fot every right question and $-1$ mark for every incorrect answer.

If the candidate attempts all of $10$ questions, then from the set $15,16,17,18,19,20$,.he can only achieve the score of 18, by marking 7 coorect options and 3 incorrect option.

If he attempts any more or less than 7 questions, then the score becomes out of the range of the set. Only if he does 7 correct questions, he can score 18 out of given set.

Now we know to get a score of 18, the candidate needs to correct only 7 questions out of 10, it means he will mark 3 questions incorrectly.

So number of possible combinations to get a score of 18 consists the number of ways he can chose the 7 correct answer multiplied by number of way he can incorrect the 3 questions.

Hence number of possible distinct combinations will be ${10}_{C_7}.(3{)}^3$

Where $(3{)}^3$ are the number of possible way he can incorrectly mark the 3 questions, as every question has 3 incorrect choice out of 4. So for 3 questions, it will $\mathrm{3.3.3}=(3{)}^3$

Hence total number of distinct possible combinations will be ${10}_{C_7}.(3{)}^3=120\times 27=3240$

So correct answer is $C$.