Math Is Fun Forum

I am confused over the wording of the problem. How is WLW 2 more? Also which team gets the home court first?

Okay, I think I've worked it out. Is it...

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Are we given that the sum of the two bad apples ≠ sum of two good apples?

It turns out that the third possibilty could be solved just in three weighing attempts, so the test ,in its three attempts and restrictions, is in fact conclusive!!!

In the third possibility, when we weigh the subgruops of group A and the subgroups of group B (using the first two attempts). We will have, in group A, one of the scale's arm be lifted and the same for group B, but we are using a scale!!, so we can know the following results:

the difference between the lighter and the heavier arms in in the first weighing attempt (weighing the subgroups of A), call it "d1", is equal to or more or less then the same difference in the second attempt (when weighing group B), call it "d2".

Now if they are equal then the two rotten apples are equal in weight to two good apples. If they are not, we cann't know which side of the balance contains the rotten apple in both weighing attempts of each group, so we remove the lighter subgroup of B from the arm (in the second attempt) and put on it the lighter side of the subgroups of A (using the third and last attempt), If the two arms balanced then we know ,for sure, that group A contains the heavier rotten apple and group B have the lighter one, So we have:

1) d1>d2 (in this case the two rotten apples weigh more then two normal ones)
2) d1<d2 ( in this case the two rotten apples weigh less then two normal)

Finally, if in the theird attempt, the scale which contains the heavier subgroup of group B remains unlifted, then we know, for sure, that group B contains the heavier rotten apple, and we have the results:

1) d2>d1 (the two rotten apples are more in weight then two normal ones)
2) d2<d1 (the two rotten are less in weight then two normal).

Note: the third possible case (the heavier subgroup of group B in the third attempt is lifted) can not happen, since the heavier rotten apple must be either in group "A" or in group "B".

Q.E.F

You don't have to know which side of the scale (in both group) contains the rotten apple!!! Why? To explain this situation suppose "without loss of generality" that group A contains the heavier rotten apple in subgroup A1, then when you weigh the subgroups of this group (group A) the arm of the scale containing the subgroup which include the heavier rotten apple (it could be A1 or A2) will lift the other arm which containing the subgroup of the normal apples and when you do the same with group B which contains the lighter rotten apple the subgroup which contains that apple (it could be B1 or B2 no matter) will be lifted, so when you cross mix the subgroups of each group,  the lifting side will match the lifted side and the rotten apples will be in the same collections.

Note: it is obvious that the thing which made you miss the point is confusing the heavier and the lighter sides with the notations of the subgrouops. I think you asked yourself the following question:
"If I donate a particular subgroup of group A "A1" and the other subgroup "A2" and do the same for group B then after weighing, the cross mix will be as follow :(A1+B2) and (A2+B1) but wait a minute!! How the hell could I know that B2 contains the lighter rotten apple or the heavier one and the same for B1?!!!" This is the mistake which you have made. We did not talk about the notations we talked about the heavier and the lighter sides!! So if group A contais  the heavier side it could be A1 or A2 no matter and the same for the lighter side it could be B1 or B2 so focus on the heavier and the lighter side not on the notations. I think it is very clear now (please let me know in any case).

Finally, if you get the third possibility, put the subgroup of the group A which weighs less with the subgroup of group B which weighs more and vice versa, and put on each arm of the scale one collection (using the third attempt), then you will get the same results that we get in the second prossibility hence you know either the two rotten apples are equal to or more or less in weight then two good apple.

Are we given that the sum of the two bad apples ≠ sum of two good apples?

anna_gg wrote:

How do you calculate first-choice v2 and v3 and also all the second choice varieties? I only understand 1st choice v1=120 (permutation of 5 numbers / 5)

Sorry, Anna, but I only have a very basic understanding of permutations, which is mainly from research I've done since you started posting permutation problems. I wasn't taught permutations at school.

anonimnystefy and others here can help you with the formulas you need.

Anyway, for what it's worth, here's how I calculated the first-choice v2 and v3 and all the second-choice varieties:

I went with what I know and like to use - an Excel spreadsheet, even though I knew it would be the long way round compared to using permutation formulas (which I did try to write, btw, but your problem had too many tricks for my basic knowledge, and my research didn't unearth the info I was looking for). 

I got Mathematica to list all 3125 (ie, 5^5) permutations (with repeat digits) of 5-digit numbers comprising the digits 1 to 5 and copied the list into an Excel spreadsheet, which I used for the rest of the project. There I whittled the list down to 2220 by eliminating permutations that contained the same digit more than twice.

I then got Excel to count the permutations that had distinct digits of specified lengths for first-choice questions:
(a) 5 distinct digits for v1 (no repeat digits)............of which there were 120    )
(b) 4 distinct digits for v2 (1 pair of repeat digits)...of which there were 1200   )  total 2220
(c) 3 distinct digits for v3 (1 pair of repeat digits)....of which there were 900    )

I could simply have deducted 120 and 1200 from 2220 to get 900, but I used the Excel count to verify that this component was going well. 

Directly underneath the row of 2220 first-choice permutations I created a row of corresponding second-choice permutations, with each one formed from leftover question numbers after taking into account those that were used by first-choice permutations. I got Excel to display them with their digits in ascending order to help with eliminating duplicates.

And that's all there was to that!