3U Maths Question Thread

[meerae]

My teacher has sent me a perms question and I haven't been able to solve it for almost two days...
The question is - how many ways can you arrange a group of people if they're put in 4 rows of 10 people each?
Can anyone help?

Thanks,
meerae

[RuiAce]

\[ \text{Make an ordered selection to place 4 out of 10 in row 1: }^{10}P_4.\\ \text{Then make an ordered selection to place 4 out of 6 in row 2: }^{6}P_4\\ \text{And finally make an ordered selection to place 2 out of 2 in row 3: }^{2}P_2 \]
\[ \text{However we haven't been told that the rows are distinguishable.}\\ \text{Therefore we would've treated arrangements of the form}\\ \boxed{ABCD}\, \boxed{EFGH}\, \boxed{IJ}\\ \text{identical to the arrangement}\\ \boxed{EFGH}\,\boxed{ABCD}\,\boxed{IJ}.\]
\[ \text{Hence to address the double counting for the two identical rows}\\ \text{an appropriate correction factor of }\frac{1}{2!}\text{ is required.}\\ \text{This gives an answer of }\frac{1}{2!} \, ^{10}P_4\, ^{6}P_4 \, ^2P_2\]
This should be identical to \( \frac{10!}{2!} \) I think

[meerae]

\[ \text{Make an ordered selection to place 4 out of 10 in row 1: }^{10}P_4.\\ \text{Then make an ordered selection to place 4 out of 6 in row 2: }^{6}P_4\\ \text{And finally make an ordered selection to place 2 out of 2 in row 3: }^{2}P_2 \]
\[ \text{However we haven't been told that the rows are distinguishable.}\\ \text{Therefore we would've treated arrangements of the form}\\ \boxed{ABCD}\, \boxed{EFGH}\, \boxed{IJ}\\ \text{identical to the arrangement}\\ \boxed{EFGH}\,\boxed{ABCD}\,\boxed{IJ}.\]
\[ \text{Hence to address the double counting for the two identical rows}\\ \text{an appropriate correction factor of }\frac{1}{2!}\text{ is required.}\\ \text{This gives an answer of }\frac{1}{2!} \, ^{10}P_4\, ^{6}P_4 \, ^2P_2\]
This should be identical to \( \frac{10!}{2!} \) I think

Thank you! But doesn't this only deal with the 10 people? when we're dealing with 40?

[RuiAce]

Thank you! But doesn't this only deal with the 10 people? when we're dealing with 40?

My bad, totally misread the question first time round.
\[ \text{It will follow a similar procedure, except now}\\ \text{we end up with picking 4 out of 40, then 4 out of 36, then 4 out of 32}\\ \text{and so on until we reach 4 out of 4.}\\ \text{This brings us to }^{40}P_4\,^{36}P_4\,^{32}P_4\, \dots ^{4}P_4\]
\[ \text{But now that I see the issue is we have 10 }\textbf{rows}\text{ and not people}\\ \text{with in fact, all 10 rows having the same size,}\\ \text{a correction factor of }\frac1{10!}\text{ is required.}\]
Final answer should now be equivalent to \( \frac{40!}{10!}\)

[terassy]

Need help please.

Prove by induction that n+(n+1)+(n+2)+...+2n=3n(n+1)/2 for all integers n ≥ 1.

I have the solutions but I don't get where they are going. For the first step, don't you just sub in 1 to 2n, but then it doesn't work with the RHS. Can someone show me how to do this question please. Cheers.

[RuiAce]

Need help please.

Prove by induction that n+(n+1)+(n+2)+...+2n=3n(n+1)/2 for all integers n ≥ 1.

I have the solutions but I don't get where they are going. For the first step, don't you just sub in 1 to 2n, but then it doesn't work with the RHS. Can someone show me how to do this question please. Cheers.

With the sub, on the LHS you just have \(1+2 = 3\). Whilst on the RHS you have \( \frac{3\times 1\times 2}{2} \) which also equals 3.

Keep in mind that \(\boxed{n=1}\) for your check. That means \(2n = 2\), and hence 2 is the number you stop at in your sum.

[terassy]

With the sub, on the LHS you just have \(1+2 = 3\). Whilst on the RHS you have \( \frac{3\times 1\times 2}{2} \) which also equals 3.

Keep in mind that \(\boxed{n=1}\) for your check. That means \(2n = 2\), and hence 2 is the number you stop at in your sum.

Wait so the sum starts with n+(n+1)?

[RuiAce]

Wait so the sum starts with n+(n+1)?

The sum starts with \(n\), and ends with \(2n\).

For \(n=1\) you have \(1+2\)
For \(n=2\) you have \(2+3+4\)
For \(n=3\) you have \(3+4+5+6\)
For \(n=4\) you have \(4+5+6+7+8\)

etc

[david.wang28]

Hello,
I'm having trouble with Question 10 in the link below. Can anyone please give me a detailed solution (I don't know where to start)? Thanks

[RuiAce]

Hello,
I'm having trouble with Question 10 in the link below. Can anyone please give me a detailed solution (I don't know where to start)? Thanks

I'm not providing the full solution to this one yet because the way of thinking follows along the lines of the angle sum question from a while back you asked. You should revisit it.

Hint: The base case should be obvious - two lines can only intersect once, or they're parallel (and thus they never intersect).

For the inductive step, use a similar train of thought. Assume that every pair of \(k\) lines intersect at most \( \frac12 k (k-1) \) times. Then, given any \(k+1\) lines, you can split them into a group of \(k\) lines (which takes advantage of the inductive assumption), and that one extra line. But at most how many times can that last line intersect with each of the \(k\) other lines?

[terassy]

Can someone check my solution for this question:
RTP: 2ⁿ > n² for all integers n ≥ 5
i.e. 2ⁿ - n² > 0

In step 2: Assume statement true for n = k
i.e. 2^k - k² > 0

Now, prove statement true for n = k+1
RTP: 2^k+1 - (k+1)² > 0

LHS = 2^k+1 - (k+1)²
        = 2^k*2 - k² - 2k - 1
        = 2(2^k - k²) + 2k² - 2k - 1          from assumption
        > 0 since (2^k - k²) > 0 from assumption and (2k² - 2k - 1) ≥ 39 since the min. value of k is 5.

Am I explaining the red part correctly?

[RuiAce]

Can someone check my solution for this question:
RTP: 2ⁿ > n² for all integers n ≥ 5
i.e. 2ⁿ - n² > 0

In step 2: Assume statement true for n = k
i.e. 2^k - k² > 0

Now, prove statement true for n = k+1
RTP: 2^k+1 - (k+1)² > 0 <- minor comment - you may want to emphasise \(k\geq 5\) here, so that later on you can shrink "since the min value of k is 5" down to "because \(k\geq 5\)"

LHS = 2^k+1 - (k+1)²
        = 2^k*2 - k² - 2k - 1
        = 2(2^k - k²) + 2k² - 2k - 1          from assumption <- note that this is not from assumption; the next line is
        > 0 since (2^k - k²) > 0 from assumption and (2k² - 2k - 1) ≥ 39 since the min. value of k is 5.

Am I explaining the red part correctly?

\[ \text{Shouldn't you get}\\ 2(2^k) - k^2 - 2k - 1 = 2(2^k - k^2) + \color{red}{k^2}-2k-1? \]
That aside, the "since \((2^k - k^2) > 0\) from assumption" bit is definitely correct. Some textbooks seem to let you go by saying something like what you did at the end, so hopefully examiners would let you go as well. I'm not too big of a fan of it though, because how do we know this for sure? What if the parabola is actually dipping (monotonic decreasing) at some point?

A workaround that I saw in the Cambridge textbook kinda gets around this issue in my opinion. It basically uses completing the square (which i know is disgusting but it works)
\begin{align*}LHS &= 2(2^k - k^2) + k^2 -2k -1\\ &> k^2 - 2k - 1\tag{from assumption}\\ &= (k-1)^2 -2\\ &> 16 - 2 \tag{since k >= 5}\\ &> 0 \end{align*}

[Jefferson]

Could someone please help me with this question.
I'm not sure where to go from
AB=5
AC=13, CD=3
BP = CQ = AE = DF = 20.

[RuiAce]

Could someone please help me with this question.
I'm not sure where to go from
AB=5
AC=13, CD=3
BP = CQ = AE = DF = 20.

(Click the image to resize it a bit better)

I omitted a lot of details here but it's basically just navigating through similar triangles. Still, feel free to mention if there's anything you want expanded on.

[Jefferson]

(Image removed from quote.)
(Click the image to resize it a bit better)

I omitted a lot of details here but it's basically just navigating through similar triangles. Still, feel free to mention if there's anything you want expanded on.

Your explanation was perfect. Thank you.
I had my head wrapped around the idea of "Pythagoras" when they banned trig (which I somehow interpreted as banning angles).
Thank you so much .