Probability (1 Viewer)

[hit patel]

A test is scheduled with 7 exams. 2 of the exams are math and 1 of the exams is English. Math exams can be scheduled consecutively.
i) how many ways can be scheduled?
ii) If english is first find prob that math is second and then other math exam is last.

Ty.

 

[BLIT2014]

@ Hit Patel is that the exact wording of the question? It seems a bit unclear/ambiguous?

Try looking at this site as it goes through probability fairly well http://www.mathsisfun.com/combinatorics/combinations-permutations.html

 

[braintic]

(i) just order the exams in 7! ways

I am assuming that the two mathS exams are considered to be 2 different exams, and not part I and part II of the same exam. In the latter case it would be 7!/2

(ii) after english, there are 6! ways of arranging the other exams. There are 2 times 4! ways of arranging them so that mathS occupies those two slots (2 ways of arranging the mathS exams in those slots, 4! ways of arranging the other exams).

So (2 times 4!) / (6!) = 1/15

 

[BLIT2014]

Part ii) is a permutation without repettion

 

[braintic]

Part ii) is a permutation without repettion

Hmmm, not sure how to interpret this. Permutations by their very nature don't allow for repetition. Otherwise we would be using powers.

 

[hit patel]

K ty

 

[hit patel]

(i) just order the exams in 7! ways

I am assuming that the two mathS exams are considered to be 2 different exams, and not part I and part II of the same exam. In the latter case it would be 7!/2

(ii) after english, there are 6! ways of arranging the other exams. There are 2 times 4! ways of arranging them so that mathS occupies those two slots (2 ways of arranging the mathS exams in those slots, 4! ways of arranging the other exams).

So (2 times 4!) / (6!) = 1/15

It says something else in answers though. well I got the same answer as u.

 

[braintic]

It says something else in answers though. well I got the same answer as u.

Could you tell me what the answers say.