Brilliant Olympiad rules

Probability Level 3

In an Olympiad exam, there are $10$ questions each having $\text{4 choices}$ out of which only 1 is correct . Question paper is divided into four sections $\text{A, B, C and D}$ containing $\text{1, 2, 3 and 4 questions respectively}$ .

In how many ways can he $\text{select and answer questions}$ if he has to score $\text{exactly 80\% marks}$ , has to attempt at least $\text{1 question}$ from each section, if it is $\underline{\text{Not}}$ necessary to $\text{attempt all questions}$ , if each question comprises of $10\text{ marks}$ and if there is $\underline{\text{No negative marking}}$ .

Bonus: Find answer if there is negative marking of -1 for each wrong attempt and this is no restriction for selecting questions.

All of my problems are original

Difficulty: $\dagger \dagger \dagger \dagger \color{grey}{\dagger}$

The answer is 683.

1 solution

Aryan Sanghi
Jan 28, 2019

There are 3 cases

Case 1:

Attempting 8 questions and getting all correct.

In this case, there are 5 options to select questions

A	B	C	D	No. Of ways
1	2	3	2	6
1	2	2	3	12
1	1	3	3	8
1	2	1	4	3
1	1	2	4	6

$\boxed{\text{Total ways }= 35}$

Case 2:

Attempting 9 questions and getting 8 correct

No. of ways of selecting 1 question and answering it wrong= No. of ways of selecting 1 question from 9 questions×3=27

In this case, there are 3 options to select questions

A	B	C	D	No. Of ways
1	2	3	3	4
1	2	2	4	3
1	1	3	4	2

$\boxed{\text{Total ways }= 27×9=243}$

Case 3:

Attempting 10 questions and getting 8 correct

No. of ways of selecting 2 question and answering it wrong= No. of ways of selecting 2 question from 10 questions×3×3=405

In this case, there is only 1 option to select questions

A	B	C	D	No. Of ways
1	2	3	4	1

$\boxed{\text{Total ways }=405×1=405}$

$\color{#3D99F6}{\boxed{\text{Total ways of scoring }80\%=35+243+405=683}}$