3U Maths Question Thread

[mirakhiralla]

Hey (:
Please help me out thank you,
Show that:
nPr= n-2Pr + (2r x n-1Pr-1)+ [r(r-1) x n-2Pr-1]

I worked on the LHS and got stuck at a certain point where my working ot looked like this:
(n-2)!/(n-r-2)! [r^2+n-1/n-r-1]

[jazzycab]

Hey (:
Please help me out thank you,
Show that:
nPr= n-2Pr + (2r x n-1Pr-1)+ [r(r-1) x n-2Pr-1]

I worked on the LHS and got stuck at a certain point where my working ot looked like this:
(n-2)!/(n-r-2)! [r^2+n-1/n-r-1]

I've played around with the right-hand side, but it seems to me that there is a sign wrong in the original question.
Is the question actually asking you to prove:

[RuiAce]

Hey (:
Please help me out thank you,
Show that:
nPr= n-2Pr + (2r x n-1Pr-1)+ [r(r-1) x n-2Pr-1]

I worked on the LHS and got stuck at a certain point where my working ot looked like this:
(n-2)!/(n-r-2)! [r^2+n-1/n-r-1]

I've played around with the right-hand side, but it seems to me that there is a sign wrong in the original question.
Is the question actually asking you to prove:

Neither of these expressions given are correct. I tried testing both of these on WolframAlpha with \( n = 10 \) and \( r = 6\). This was what I got.

LHS
Given RHS
Proposed RHS (Note: I think the answer of 0 was a coincidence.)

Still, the original question is wrong.

[mirakhiralla]

I've played around with the right-hand side, but it seems to me that there is a sign wrong in the original question.
Is the question actually asking you to prove:

Yeah, last one is wrong, sorry. It's supposed to be n-2Pr-2,

so the correct one is:

nPr= n-2Pr + (2r x n-2Pr-1) + [r(r-1) x n-2Pr-2]

[RuiAce]

Yeah, last one is wrong, sorry. It's supposed to be n-2Pr-2,

so the correct one is:

nPr= n-2Pr + (2r x n-2Pr-1) + [r(r-1) x n-2Pr-2]

\begin{align*}RHS&= ^{n-2}P_r + 2r ^{n-2}P_{r-1} + r(r-1) ^{n-2}P_{r-2}\\ &= \frac{(n-2)!}{(n-2-r)!} + \frac{2r(n-2)!}{(n-r-1)!} + \frac{r(r-1)(n-2)!}{(n-r)!}\\ &= \frac{(n-r)(n-1-r)(n-2)!}{(n-r)!} + \frac{2r(n-r)(n-2)!}{(n-r)!} + \frac{r(r-1)(n-2)!}{(n-r)!}\\ &= \frac{(n-2)!}{(n-r)!} \left[ (n-r)(n-r-1) + 2r(n-r) + r(r-1)\right]\\ &= \frac{(n-2)!}{(n-r)!}(n^2 - nr - n - nr + r^2 + r + 2nr - 2r^2 + r^2 - r)\\ &= \frac{(n-2)!}{(n-r)!}(n^2 - n)\\ &= \frac{n(n-1)(n-2)!}{(n-r)!}\\ &= \frac{n!}{(n-r)!}\\ &= ^nP_r = LHS\end{align*}

Working on the LHS first when the RHS is heaps messier isn't really a good idea.

[Dragomistress]

In how many ways can fie letters be chosen from the letters ARRANGE?

[arii]

In how many ways can fie letters be chosen from the letters ARRANGE?

There are 7 letters in total. For the letter repetitions, there is 2 of A, 2 of R.

Therefore, total ways is (7!)/(2!x2!) = 1260.

EDIT: Refer to Rui's solutions. Whilst skim reading, I didn't notice the question wanted FIVE letters as the asker incorrectly spelt it.

[RuiAce]

There are 7 letters in total. For the letter repetitions, there is 2 of A, 2 of R.

Therefore, total ways is (7!)/(2!x2!) = 1260.

Careful. I think she wanted only 5 out of the 7 letters.

In how many ways can fie letters be chosen from the letters ARRANGE?

If I've misinterpreted then just go with what arii said. Otherwise, confirm for me that you only want 5 letters to be chosen and not necessarily arranged?

[Dragomistress]

I don't quite get what you mean, but the question asked,
In how many ways can five letters be chosen from the letters ARRANGE?

I think, I want 5 letters to be chosen from ARRANGE where order matters.

[RuiAce]

I don't quite get what you mean, but the question asked,
In how many ways can five letters be chosen from the letters ARRANGE?

I think, I want 5 letters to be chosen from ARRANGE where order matters.

Problems in perms and combs are split into selections and arrangements.

Selections are divided into ordered and unordered, as well as with or without repetition. Arrangements are just determined with or without repetition. (See last year's 4U trial survival lectures for more information.) Sometimes, you only need to select people (e.g. four out of ten to have a meeting), but then you may need to arrange them (e.g. put those four people in a circle).
________________________________________________

The cases for this question are:
1. Two A's and two R's appear
2. One A's and two R's appear
3. Two A's and one R's appear
4. Two R's appear but no A's
5. Two A's appear but no R's
6. One of both A and R appear
(Not that any further cases are impossible, since we need a total of 5 letters)

If all we needed to do was to choose five letters, which actually implies unordering, then there can only be \(3\) ways of attaining cases 1, 2 and 3, whilst there are \(6\) ways of attaining cases 4, 5 and 6. (This is because for cases 1, 2 and 3, we need to pick one more letter from {N,G,E}. Whereas for cases 4, 5 and 6 we need to pick two more from that set.)

Whereas if we had to arrange them as well, we would need more work.
Case 1: Multiply \(3\) to \( \frac{5!}{2!2!} \)
Case 2: Multiply \(3\) to \( \frac{5!}{2!} \)
Case 3: As with case 2
Case 4: Multiply \(6\) to \frac{5!}{2!} \)
Case 5: As with case 4
Case 6: Multiply \(6\) to \(5!\)
Note that we have multiplied the number of ways we can choose the letters, to the number of ways we can then arrange them to make words.

[bobcheng1111]

Hi! I was wandering how to do this question:

The coefficients of x^4 and x^5 in the expansion of (3-x)^n are equal in magnitude but opposite in sign. Find the value of N

[RuiAce]

Hi! I was wandering how to do this question:

The coefficients of x^4 and x^5 in the expansion of (3-x)^n are equal in magnitude but opposite in sign. Find the value of N

[bobcheng1111]

Hey! I can understand up to the step right after you expanded  3(n 4) = (n 5), but I was wandering how you simplified the expansion. Cheers

[RuiAce]

Hey! I can understand up to the step right after you expanded  3(n 4) = (n 5), but I was wandering how you simplified the expansion. Cheers

Going from \( 3 \times \frac{n!}{4!(n-4)!} = \frac{n!}{5!(n-5)!} \) to \( 3\times \frac{5!}{4!} = \frac{(n-4)!}{(n-5)!} \) was just cancelling out the \(n!\)'s and then cross multiplying the \(5!\) and \((n-4)!\).

And then as said, after that I used the formula \( r! = r(r-1)!\) You may or may not not have seen it before, but if you haven't you should understand why it works.

[bobcheng1111]

Going from \( 3 \times \frac{n!}{4!(n-4)!} = \frac{n!}{5!(n-5)!} \) to \( 3\times \frac{5!}{4!} = \frac{(n-4)!}{(n-5)!} \) was just cancelling out the \(n!\)'s and then cross multiplying the \(5!\) and \((n-4)!\).

And then as said, after that I used the formula \( r! = r(r-1)!\) You may or may not not have seen it before, but if you haven't you should understand why it works.

ohhh I think I get it now thank you !