HW 15 1) Here is a sample proof that contains an error. Explain why the proof is

Question

HW 15 1) Here is a sample proof that contains an error. Explain why the proof is not correct. Theorem: R is a relation on a set A that is symmetric and transitive. Then R is reflexive. Proof: Let R be a relation on a set A that is symmetric and transitive. Let aEA . Take be A such that (a,b) e R. Because R is symmetric, (b,a) e R. Now using the transitive property, we can conclude that (a,a)e R. QED 2) A market sells 40 kinds of candy bars. You want to buy 20 candy bars. a) How many possibilities are there if you want at most four toffee bars, at most six mint bars, at least three peanut butter bars but at most six peanut bars.

Explanation / Answer

Hi,
1. The proof is wrong because we are assuming something that may not be true,
we are assuming a 'b' exists such that (a,b) is in R, but that is not true overall, thats why the proof is wrong.
Take the example of A={0,1,2}
R={(0,1),(1,0),(0,0),(1,1)}
we can see that if we chose a as 2 here, there is no b such that (a,b) is in R, thats why the above relation os symmetric and transitive but not reflexive
2.
Given there are total 40 kinds of candy bars and we want to buy 20 candy bars
therfore the total possibilities are 59C39
now, we have few conditions, they are
1. at most 4 toffeebars
2. at most 6 mintbars
3. at least 3 but atmost 6 peanut butter bars
lets incorporate these conditions
to calculate the atmost, we just remove the compliment from total i.e
to calculate number of atmost 4 toffee bars we remove number of possibilities of 5 toffebars from total
similarly with mint bars and peanut bars
that is 59C39 - (54C39)-(52C39)-(52C39) + 50C39 (for choosing the remaning 5+7+7 bars)
to get atleast 3 peanut butter bars 51C39
Therefore total is 59C39 - (54C39)-(52C39)-(52C39) + 50C39 + 51C39

Thumbs up if this was helpful otherwise let me know in comments

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