Probability Problem, permutation or combination
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Probability Problem, permutation or combination

[From: ] [author: ] [Date: 11-11-01] [Hit: ]
among others:{(1,2), (2,1), (3,18),......

C = 326! / [2!(326 - 2)!] = 326! / (2!*324!) = (326*325*324!) / (2!*324!) = 326*325 / 2! =

= 326*325 / 2 = 52975

Your answer is 52975 handshakes.

Your second question is a bit different because in the groups of answers we would count, for example, student 3 giving a treat to student 18 AND student 18 giving a treat to student 3. So in the list of groups we would have, among others:

{(1,2), (2,1), (3,18), (18,3), (54,187), (187,54),... etc.}

So in this case the order DOES matter, so it's a permutation case, and again there is no repetition because a student cannot give himself or herself a treat, so groups like (7,7) or (278,278) do not count. The formula for this case is:

m! / (m - n)!

m = total number of elements, which is 326
n = total number of elements per group, which is 2

m! / (m - n)! = 326! / (326 - 2)! = 326*325*324! / 324! = 326*325 = 105950

That's your answer: 105950 treats exchanged.

I hope this helps!

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Permutation is when order matters. Combination is when order does not matter. In this scenario, order does not matter, so it would be a combination.

And no, the answer would not be the same. If order matters (i.e. a permutation), then ABCD is not the same as DCBA. With a combination, they are equal, thus there will be much less solutions to a combination.

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1)326C2=52975
Combination is used,
1st student shakes with 325
2nd with ... 324
and thus on.

2)326*325=105950
you can imagine this one

Hope that helps!
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keywords: combination,Problem,or,permutation,Probability,Probability Problem, permutation or combination
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