4U Maths Question Thread

[RuiAce]

Hi znaser,

I might be wrong, but I'm pretty sure that with your method, for the white flowers; it's just 6x3! as if you imagine the 3 flowers as a bunch, you can order them with only 6 choices and not 6! choices.

So instead, the answer should be 8!-3!x6= 40284, unless if I have done something horribly wrong

Judging by your 6*3!, you probably forgot that all the red roses are different as well.

That's my guess.

[znaser]

I follow you! That covers cases with all three white roses together, but if you think carefully, it actually doesn't cover two at a time. So, you aren't subtracting enough cases, which is why your answer is higher than it should be

I love the method you use, and often it works really well, just not in this case, you need to be very careful with probability!

Thanks! I think i follow now. So that method would also include options like WRWWRRRR where 1 white rose may be separated but two may be together.

[znaser]

Hi znaser,

I might be wrong, but I'm pretty sure that with your method, for the white flowers; it's just 6x3! as if you imagine the 3 flowers as a bunch, you can order them with only 6 choices and not 6! choices.

So instead, the answer should be 8!-3!x6= 40284, unless if I have done something horribly wrong

Hi Mei2016,
I think it would be 8! - 6! x 6 x 3! as that would remove the possibility for 1 white rose to be separated but 2 to be together. You get the answer but it may be a fluke 😅. But yh either way rui's method is probably better in this case as it's much more simpler.

[Ali_Abbas]

Can't exactly follow your logic how 6!3! eliminates the cases when the white roses are planted together in your working.

I'm no expert in permutations and combinations, but I believe Rui's answer is incorrect. To demonstrate this, we provide the following counterexample:

Suppose the red roses are fixed as Rui imposed. Then observe that another viable arrangement is:

_RRRR_R_

Here, there are only three blank spots to fill. Thus, Rui's calculation does not capture the entire spectrum of cases and I conjecture that to achieve the correct number of permutations using this method, one would have to apply brute force to obtain all possible arrangements of red roses that leave at most one blank spot in between any two that occur. This would prove to be an inefficient process.

I believe one approach that seems right, though I'm not entirely convinced that it is, is to group two white roses together, treating them as a single object, and evaluate the number of permutations this yields, then subtracting this off from the total number of arrangements, namely 8!. The important observation to make is that this will account for the case where all three white roses are together, inclusively. So we need not treat the two cases separately as requiring two independent calculations.

When two white roses are grouped together, we now have 7 distinct objects. These are arranged in 7! ways, however, the two white roses that are coupled can be chosen in ³P₂ = 3! ways. So we thus have 7!3! ways this can happen.

The total number of arrangements where no two white roses are together is calculated as 8! - 7!3! = 10 080. However, this is lower than Rui's answer which I alluded to have understated the true answer since he only accounted for one special arrangement of the red roses. Hopefully, I still have the right idea in my method but just need to make a small alteration. Feel free to correct me where I have gone wrong.

[jamonwindeyer]

I'm no expert in permutations and combinations, but I believe Rui's answer is incorrect. To demonstrate this, we provide the following counterexample:

Suppose the red roses are fixed as Rui imposed. Then observe that another viable arrangement is:

_RRRR_R_

Here, there are only three blank spots to fill. Thus, Rui's calculation does not capture the entire spectrum of cases and I conjecture that to achieve the correct number of permutations using this method, one would have to apply brute force to obtain all possible arrangements of red roses that leave at most one blank spot in between any two that occur. This would prove to be an inefficient process.

I believe one approach that seems right, though I'm not entirely convinced that it is, is to group two white roses together, treating them as a single object, and evaluate the number of permutations this yields, then subtracting this off from the total number of arrangements, namely 8!. The important observation to make is that this will account for the case where all three white roses are together, inclusively. So we need not treat the two cases separately as requiring two independent calculations.

When two white roses are grouped together, we now have 7 distinct objects. These are arranged in 7! ways, however, the two white roses that are coupled can be chosen in ³P₂ = 3! ways. So we thus have 7!3! ways this can happen.

The total number of arrangements where no two white roses are together is calculated as 8! - 7!3! = 10 080. However, this is lower than Rui's answer which I alluded to have understated the true answer since he only accounted for one special arrangement of the red roses. Hopefully, I still have the right idea in my method but just need to make a small alteration. Feel free to correct me where I have gone wrong.

I believe that Rui's answer accounts for that; he has all possible places for the white roses to be placed. Taking his example:



We would put white roses in the first, fifth, and last positions to obtain your arrangement

[Ali_Abbas]

I believe that Rui's answer accounts for that; he has all possible places for the white roses to be placed. Taking his example:



We would put white roses in the first, fifth, and last positions to obtain your arrangement

Thanks Jamon I totally missed this fact before.

I believe I know what was wrong with my method (with some help from Rui). By pairing two of the white flowers and calculating the number of corresponding arrangements, I seemed to have double-counted the number of times where all three white flowers are together and hence, subtracted two multiples of the same thing giving a deficient answer. To understand why, we offer the following explanation:

Suppose we fix a pair of white flowers. Observe that no matter where this pair occurs, there is at least one neighbouring spot to be filled. This neighbouring spot has the option of being filled by the remaining white flower. When this occurs, we also have the option of interchanging the three whites, given that they are distinct. As we shift the paired flowers across one space at a time, the property of having at least one neighbouring spot is preserved. In particular, translating the pair across by a single space, we observe that all previous arrangements of the three whites occurring together can be reproduced simply by inputting the third white flower on the opposite side of the paired group to which it previously occupied, and again allowing all three to be interchanged. Visually, this can be represented as follows:

....(WW)W.... (this is where the third white is to the right of the pair of whites and before we shift them across)

--> ....W(WW).... (this is after translating the pair horizontally to the right by one space and inputting the third white on the opposite side, namely the left)

Clearly, an overlapping of arrangements occurs. We can be sure that each arrangement occurs exactly twice as applying a second translation in the same direction as the first, either the left-most or right-most space (depending on which direction we are traversing across) becomes excluded from the pact and a new space joins instead (to be made up of all three whites).

Thus, to maintain the correct number of permutations we desire, we must add a single multiple of the number of permutations of when all three flowers are together. This is easily calculated as 6!3!, as stated by other members.

Hence, the correct number of permutations using my method is:

8! - 7!3! + 6!3! = 14 400, as per Rui's answer.

QED

 

[RuiAce]

Ah so it does involve inclusion-exclusion.

[Ali_Abbas]

Ah so it does involve inclusion-exclusion.

I had to think about it for a while but yes, that was the underlying issue.

[Mei2016]

Hi Mei2016,
I think it would be 8! - 6! x 6 x 3! as that would remove the possibility for 1 white rose to be separated but 2 to be together. You get the answer but it may be a fluke 😅. But yh either way rui's method is probably better in this case as it's much more simpler.

Hey znaser,

You mean I got the answer? Do you know what it is? Also, where did this question come from? (as it would normally be like 5 girls and 3 boys...)
I really like Rui's way of doing it, so if you have the answers, what source is if from?

Also, to continue with my method, (because I had previously neglected the white flowers being together case), I've come to:

All possibilities-all white tog-considering the case of 2 white flowers tog and one separate;
8!-6!x6x3!-5!x6x5x2= 7200 (which is exactly half of Rui's answer, and that's so interesting, but I don't know what's wrong with it)

[Mei2016]

Hi,

This is Q6 iv) from the 1997 HSC paper. I just don't get the N.B part in the answers.
(i.e how the _< turned into a < sign)

[znaser]

Hey znaser,

You mean I got the answer? Do you know what it is? Also, where did this question come from? (as it would normally be like 5 girls and 3 boys...)
I really like Rui's way of doing it, so if you have the answers, what source is if from?

Also, to continue with my method, (because I had previously neglected the white flowers being together case), I've come to:

All possibilities-all white tog-considering the case of 2 white flowers tog and one separate;
8!-6!x6x3!-5!x6x5x2= 7200 (which is exactly half of Rui's answer, and that's so interesting, but I don't know what's wrong with it)

The question is from an S&G paper back in 2013. It's multiple choice and the answer is 14 400 (the same as Rui's). They did it the same way as him but their explanation was a bit confusing. Not sure if u've heard of their company but they're known for their rlly difficult papers relative to others.

[RuiAce]

Hi,

This is Q6 iv) from the 1997 HSC paper. I just don't get the N.B part in the answers.
(i.e how the _< turned into a < sign)

[Ali_Abbas]

Hi,

This is Q6 iv) from the 1997 HSC paper. I just don't get the N.B part in the answers.
(i.e how the _< turned into a < sign)

As per the explanation given in the set of solutions, the key observation they have made is that the upper and lower bounds are irrational numbers, given that the number Pi is present in both cases. The intermediately-occurring sum of rationals will, by intuition, evaluate to another rational. The set of rational numbers and the set of irrational numbers are disjoint sets. That is, they do not have any common elements and so this facilitates their justification for replacing the "less than or equal to" symbols with strict inequalities as equality cannot hold between numbers that belong to non-overlapping sets.

EDIT: In case you are unsure about my wording, I will define some of the terms used.

Upper and lower bounds are the numbers that occur on the "outside" of the inequality statement. In this case, the lower bound is Pi/4, and the upper bound is Pi/4 + 1/1003.

An intermediate term is one which occurs in between, or roughly speaking, in the middle of the string of terms in the statement we are considering.

Again, disjoint simply means non-overlapping or mutually exclusive. Basically it's where nothing is common.

Strict inequalities are < and >.

Hopefully that alleviates any potential confusion.

[Mei2016]

Hi Rui,

Thanks for your help. I know what a rational number is, I was just previously confused with the wording of their explanation.
In the test when we do this (taking the equality out) do we need to explain it in words like how the book does?

[RuiAce]

Hi Rui,

Thanks for your help. I know what a rational number is, I was just previously confused with the wording of their explanation.
In the test when we do this (taking the equality out) do we need to explain it in words like how the book does?

If you're going to make an inequality strict you have to be explicit about why.

For something a bit more trivial like 1≤2 to 1<2 you can easily argue this by saying "because 1≠2".

Since what your given is far more complicated and the absence of equality is less obvious, you have to say something.


Your explanation doesn't have to be identical to the book's if that's your query. It just has to allow your logic and reasoning to flow.