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CS4963
HW1 - Does order matter?

Author	Message
wveit Posts: 18	Posted 16:46 Feb 01, 2018 \| In Assignment A1, I have a question about section IV question 1. The question asks: Set-1 and Set-2 are two sets consisting of 10 and 12 elements respectively. In how many ways can 5 elements be selected from Set-1? When the question asks "how many ways", I'm trying to figure out if the order in which the 5 elements are selected matters or not. It seems like this question can be interpreted 2 ways: One interpretation: It is asking how many different groups of 5 can be selected from Set-1. This means that order in which the 5 elements are picked doesn't matter. Other interpretation: It is asking how many different sequences of 5 can be selected from Set-1. This means that the order in which the 5 elements are picked does matter. Actually, section IV questions 1, 2, and 3 have the same ambiguity. Is it possible to have some clarification if order is supposed to matter for these problems? Thank you!
ChristopherNg Posts: 17	Posted 16:59 Feb 01, 2018 \| I also was wondering that. Also when combining the two sets, are we assuming no two elements are the same?
marrawi Posts: 21	Posted 17:05 Feb 01, 2018 \| This pertains to Combination formula. C(n, r) or nCr = n!/r!(n−r)!
marrawi Posts: 21	Posted 17:17 Feb 01, 2018 \| As for 2 and 3: Rule of Product - Statement: If there are n ways of doing something, and m ways of doing another thing after that, then there are n x m ways to perform both of these actions. Rule of Sum - Statement: If there are n choices for one action, and m choices for another action and the two actions cannot be done at the same time, then there are n+m ways to choose one of these actions. Disclaimer: This is what I remember from discrete structures class I took years ago.
wveit Posts: 18	Posted 19:01 Feb 01, 2018 \| Thanks for your quick replies! Since you mentioned the Combination formula, you must believe the "order does not matter" interpretation. Is there a reason you chose that interpretation and not the other? I'm trying to see if that is the same interpretation that our instructor will have (or that the exam will have, if we get a question with that wording). For now I'll stick with using combination formula instead of permutation formula. Thanks again
marrawi Posts: 21	Posted 19:23 Feb 01, 2018 \| Hmmmm, not sure how you would pick permutation for this case. Even if we consider it, permutation formula is P ( n , r ) = n ! ( n − r ) ! => P ( 10 , 2 ) = 10!(10-2)! = 90. I don't see any choice that you can select for this answer.
wveit Posts: 18	Posted 19:26 Feb 01, 2018 \| None of the above is an answer
wveit Posts: 18	Posted 19:30 Feb 01, 2018 \| Also in question 1, we are selecting 5 items from a set of 10. If we use permutation formula ₁₀P₅ = 10! / (10 - 5)! = 10! / 5! . That is one of the answers. Last edited by wveit at 19:31 Feb 01, 2018.
marrawi Posts: 21	Posted 19:42 Feb 01, 2018 \| Sorry, I thought it was 2 elements. Last edited by marrawi at 19:43 Feb 01, 2018.
wveit Posts: 18	Posted 19:53 Feb 01, 2018 \| No problem at all. That is how our discussions about these problems have been going all week.
BrianK Posts: 25	Posted 20:48 Feb 01, 2018 \| Order does not matter. Later on the questions ask for "in a line", so I'm assuming thats where order matters. If it's asking just to select 5 random elements from set 1, then it's probably a combination problem. I just went with ₁₀C₅
rpamula Posts: 225	Posted 07:30 Feb 02, 2018 \| Yes. It is a "combination" problem if order does not matter. I see some ambiguity in the way Problem #2 is stated. I will add a clause at the end to say if "Set1 and Set2 are put together"
marrawi Posts: 21	Posted 09:15 Feb 02, 2018 \| I see order matters on Q11: "round robin fashion"
rpamula Posts: 225	Posted 09:22 Feb 02, 2018 \| It is not "order" in terms of "permutations". There is match between "any" two participants. Find out how many matches occur. The problem appears tricky as there is a second go around.
marrawi Posts: 21	Posted 09:43 Feb 02, 2018 \| rpamula wrote: It is not "order" in terms of "permutations". There is match between "any" two participants. Find out how many matches occur. The problem appears tricky as there is a second go around. Aha! Although the # would be the same (just in this case), your logic is the correct one. Thank you!