Communication
November 28, 2024STQC-NIELIT SC-B 2021
| Question 30 |
The students in a class are seated according to their marks in the previous examination. Once it so happens that four of these students get equal marks and therefore the same rank. To decide their seating arrangement, the teacher wants to write down all possible arrangements, one in each separate bit of paper, in order to choose one of these by lots. How many bits of paper are required?
| 9 | |
| 12 | |
| 15 | |
| 24 |
Question 30 Explanation:
We are given that four students got equal marks.
On one bit of paper, one arrangement of rank is to be written.
Let the students be named as A,B,C and D.
Now, A can be treated as having rank I in 4 ways.
B can be treated as having rank II in 3 ways.
C can be treated as having rank III in 2 ways.
D can be treated as having rank IV in 1 way.
Therefore, total number of bits of paper required for all arrangement
=4×3×2×1=24
On one bit of paper, one arrangement of rank is to be written.
Let the students be named as A,B,C and D.
Now, A can be treated as having rank I in 4 ways.
B can be treated as having rank II in 3 ways.
C can be treated as having rank III in 2 ways.
D can be treated as having rank IV in 1 way.
Therefore, total number of bits of paper required for all arrangement
=4×3×2×1=24
Correct Answer: D
Question 30 Explanation:
We are given that four students got equal marks.
On one bit of paper, one arrangement of rank is to be written.
Let the students be named as A,B,C and D.
Now, A can be treated as having rank I in 4 ways.
B can be treated as having rank II in 3 ways.
C can be treated as having rank III in 2 ways.
D can be treated as having rank IV in 1 way.
Therefore, total number of bits of paper required for all arrangement
=4×3×2×1=24
On one bit of paper, one arrangement of rank is to be written.
Let the students be named as A,B,C and D.
Now, A can be treated as having rank I in 4 ways.
B can be treated as having rank II in 3 ways.
C can be treated as having rank III in 2 ways.
D can be treated as having rank IV in 1 way.
Therefore, total number of bits of paper required for all arrangement
=4×3×2×1=24
Subscribe
Login
0 Comments