Group Exam

Let in ABC school there are two groups X and Y.The two groups will give a 4 days long exam consisting 4 subjects.On a day two groups will give this exam simultaneously.But on a day two groups will give exam of different subjects,this means on a day two groups can not give the test on single subject.e.g.on the first day both the groups can not give a test on mathematics but one can give physics and other math.Then the question is - In how many ways the exam shedule can be arranged?


The answer is 216.

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2 solutions

Alapan Das
Dec 13, 2018

Its certain that if we take X as fundamental or primary then Y depends on X.Then X is independent of Y.Now X group can give the 4 days long exam in 4!=24 ways .And for every such schedule Y can give the exam in 9 ways.Actually schedule of Y is the disarrangement of the schedule of X.So the number of arrangement of Y against X is 4!(1/2!-1/3!+1/4!)=9.Then the total number of arrangement is 24*9=216

Lucas Tan
Dec 19, 2018

Let Y be dependent on X, and X be independent.

Number of free ways to arrange X = 4 ! = 4!

Consider the 4 subjects ABCD in this specific order.

Number of ways for any 1 subject to be fixed = 4 × 3 ! = 24 = 4 \times 3! = 24

Number of ways for any 2 subjects to be fixed = 6 × 2 ! = 12 = 6 \times 2! = 12

Number of ways for any 3 subjects to be fixed = 3 × 1 ! = 3 = 3 \times 1! = 3

By Principles of Inclusion and Exclusion, Number of ways Y cannot be = 24 12 + 3 = 15 = 24 -12 + 3 = 15

\therefore Number of ways Y can be = 4 ! 15 = 9 = 4! - 15 = 9

Hence, Number of ways to arrange X and Y = 4 ! × 9 = 216 = 4! \times 9 = 216

What you have done that is actually dis arrangement,what I have written above .Your explation is perfect and proves the disarrengment formula.

Alapan Das - 2 years, 5 months ago

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Ohh, I didn't really know what was disarrangement haha, sorry about that. Could you explain to me the derangement formula? I don't exactly understand how it works.

Lucas Tan - 2 years, 5 months ago

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