Don't understand this problem (Permutations, I think)

[cupcakes]

Homework Statement

Please see attached image.

Homework Equations

The Attempt at a Solution

I don't understand the problem at all. Can someone explain to me what the problem is stating, more precisely what an arrangement is? An example would be nice. Thanks.

 

[Ray Vickson]

The note at the bottom explains exactly way is meant by an arrangement. What part of the explanation is unclear to you?

RGV

 

[cupcakes]

So let's say instead of 2 to 2012 its from 7 to 12. would one possible arrangement be {10,12,9,7,8,11} → 101297811 ? if that's so I'm guessing there are much more possible arrangements of 2 to 2012 than 2012. Yet the product only has up to a₂₀₁₂. So is this some sort of probability question? Thanks.

 

[Ray Vickson]

So let's say instead of 2 to 2012 its from 7 to 12. would one possible arrangement be {10,12,9,7,8,11} → 101297811 ? if that's so I'm guessing there are much more possible arrangements of 2 to 2012 than 2012. Yet the product only has up to a₂₀₁₂. So is this some sort of probability question? Thanks.

Of course there are tons of different arrangements. Now go back and read what you are asked in parts (a), (b) and (c). These talk about "every" arrangement in (a) and (b), and "some arrangements" in (c).

Where do you see anything about "probability"? The question never said anything about randomness, or chance, or whatever.

RGV

 

[Chestermiller]

Homework Statement

Please see attached image.

Homework Equations

The Attempt at a Solution

I don't understand the problem at all. Can someone explain to me what the problem is stating, more precisely what an arrangement is? An example would be nice. Thanks.

This is not a problem in permutations. There are 1006 even numbers between 2 and 2012, and there are 1005 odd numbers between 3 and 2011. You are trying to determine whether the product of the factors in your example is even or odd when you substitute each one of the 2011 even and odd integers into the 2011 available variable slots in your product expression. If you pair the 1006 even integers with the 1006 even integers already present in the factors, and you pair the 1005 odd integers with the 1005 odd integers already present in the factors, you, of course end up with an even product. See what happens when you try to pair the 1005 odd integers with 1005 of the 1006 even integers already present in the factors.

 

[cupcakes]

This is not a problem in permutations. There are 1006 even numbers between 2 and 2012, and there are 1005 odd numbers between 3 and 2011. You are trying to determine whether the product of the factors in your example is even or odd when you substitute each one of the 2011 even and odd integers into the 2011 available variable slots in your product expression. If you pair the 1006 even integers with the 1006 even integers already present in the factors, and you pair the 1005 odd integers with the 1005 odd integers already present in the factors, you, of course end up with an even product. See what happens when you try to pair the 1005 odd integers with 1005 of the 1006 even integers already present in the factors.

From what Ray said, I gathered that we substitute arrangements of the 2011 numbers. However, if I understand you correctly, you're saying that we substitute the integers themselves. I'm confused...

Edit

An attempt at the problem:

ODD + ODD = EVEN
ODD + EVEN = ODD
EVEN + EVEN = EVEN
--------------------------
ODD*ODD = ODD
ODD*EVEN = EVEN
EVEN*EVEN = EVEN
--------------------------

So the only way the product will be odd is if each factor is odd. So if the odd arrangements are matched with the factors with even numbers and the even arrangements with the factors with odd numbers, the product will be odd. So we've eliminated the possibility of (a) as an answer, right?

Now if we match the even arrangements with factors with even numbers and odd arrangements with factors with odd numbers, each factor will be even. Consequently, the product will be even.

So that would make the answer (c). Am I correct? Thanks.

 

[Ray Vickson]

Sigh! The problem is perfectly clear, but for some reason you are just not getting it. Let me do a smaller example, where I use numbers 2,3,4,5 instead of 2,3,4,...,2012. In the little example, if a2,a3,a4,a5 is an arrangement (= permutation of 2,3,4,5) the product is P = (2+a2)*(3+a3)*(4+a4)*(5+a5). For example, for arrangement 5,4,3,2 we get P = (2+5)*(3+4)*(4+3)*(5+2). For arrangement 4,2,5,3 we get P = (2+4)*(3+2)*(4+5)*(5+3), etc. So, for different arrangements (i.e., different permutations) we get different values of P. There are 24 different arrangements, so there could be as many as 24 corresponding values of P.

The same type of thing happens in the larger problem. Basically, you want to know if all values of P are even, or if all are odd, or if there are some that are even and some that are odd. However there are now about 0.7*10^(5772) arrangements, which is much, much larger than the total number of electrons in the whole universe, so you cannot answer the question by just listing all the values.

RGV

 

[Chestermiller]

From what Ray said, I gathered that we substitute arrangements of the 2011 numbers. However, if I understand you correctly, you're saying that we substitute the integers themselves. I'm confused...

Edit

An attempt at the problem:

ODD + ODD = EVEN
ODD + EVEN = ODD
EVEN + EVEN = EVEN
--------------------------
ODD*ODD = ODD
ODD*EVEN = EVEN
EVEN*EVEN = EVEN
--------------------------

So the only way the product will be odd is if each factor is odd. So if the odd arrangements are matched with the factors with even numbers and the even arrangements with the factors with odd numbers, the product will be odd. So we've eliminated the possibility of (a) as an answer, right?

Now if we match the even arrangements with factors with even numbers and odd arrangements with factors with odd numbers, each factor will be even. Consequently, the product will be even.

So that would make the answer (c). Am I correct? Thanks.

You're on the right track. You almost solved it, but not quite. This is because there is one more even integer than odd integer (1006 vs 1005). So if you try to make the product odd, by matching all the odd numbers with the even integers in the factors, and the even arrangements with the factors with odd numbers, there will be one even integer and one even number in a factor left over. These will sum to an even number, so the product will have one even factor, and will be even. It is impossible to make the product odd.

 

[cupcakes]

Sigh! The problem is perfectly clear, but for some reason you are just not getting it. Let me do a smaller example, where I use numbers 2,3,4,5 instead of 2,3,4,...,2012. In the little example, if a2,a3,a4,a5 is an arrangement (= permutation of 2,3,4,5) the product is P = (2+a2)*(3+a3)*(4+a4)*(5+a5). For example, for arrangement 5,4,3,2 we get P = (2+5)*(3+4)*(4+3)*(5+2). For arrangement 4,2,5,3 we get P = (2+4)*(3+2)*(4+5)*(5+3), etc. So, for different arrangements (i.e., different permutations) we get different values of P. There are 6 different arrangements, so there could be as many as 6 corresponding values of P.
RGV

Thank you, I finally understand what it means. It may be easy for you but I've never learned permutations or anything like this. This problem is not course material, just something I came across.

You're on the right track. You almost solved it, but not quite. This is because there is one more even integer than odd integer (1006 vs 1005). So if you try to make the product odd, by matching all the odd numbers with the even integers in the factors, and the even arrangements with the factors with odd numbers, there will be one even integer and one even number in a factor left over. These will sum to an even number, so the product will have one even factor, and will be even. It is impossible to make the product odd.

Alright that makes sense. Since we can prove that there will always be at least one even factor, the product will always be even. So the answer is (a). Thanks Chestermiller.