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Re: HSC 2015 3U Marathon NEW QUESTION: $ Six diners in a restaurant choose randomly from a menu featuring six main courses. Find the probability that exactly one of the main courses is NOT chosen by any of the diners. $

Re: HSC 2015 3U Marathon NEXT QUESTION: $ Find the number of ways in which the letters of the word PERMUTATION are arranged such that the vowels occupy themselves in the order AEIOU from left to right, but not necessarily together. $

Re: HSC 2015 3U Marathon .

Re: HSC 2015 3U Marathon So you will have 6 numbers beginning with 4, 6 with 3, 6 with 2, and 6 with 1, so the sum of all of these is 24000 + 18000 + 12000 + 6000 = 60000 (considering only the thousands place in each number). Then you have the same thing for the hundreds, tens and ones place...

Re: HSC 2015 4U Marathon Yes, that should be fine too, as long as you can prove those other expressions are imaginary, the proof is valid.

Re: HSC 2015 4U Marathon $ Since the diagonals of a rhombus are perpendicular then $z+w=ki(z-w)$ \Rightarrow \, $\frac{z+w}{z-w}=ki$ \Rightarrow \, $\arg(\frac{z+w}{z-w}) = \pm\frac{\pi}{2}$ for all real $k$. $

Re: HSC 2015 4U Marathon $ Since the moduli of $z$ and $w$ are equal, the parallelogram formed using these vectors actually becomes a rhombus. So the diagonals (given by $z+w$ and $z-w$) are perpendicular to each other, and hence the expression holds true. $

OK, so you will have 6 mini-quizzes regarding MATLAB and you actually don't have to learn MATLAB package properly at all. They are really simple and you will be given multiple attempts (I think 5) for each quiz. There should be self-paced lessons via Moodle and you can use that to learn basics...

ACCT1501: Textbook is useless, go into the lectures, absorb/learn everything you can, go home, read over lecture notes and try to understand as much as possible, if still hard, then use Google/YouTube, if you still don't get it, ask your tutors in your tutorials. Do all your tutorial homework...

Again, you apply the same procedure of solving the two equations simultaneously to obtain points of intersection which will then determine your limits of integration. So you would to 3(2x - x^2) = x^2 - 4x + 6 --> x = 3/2 or x = 1; then integrate over your desired interval.

$The integral would be given as $\int_{0}^{1} x^{\frac{1}{2}}-x^{\frac{3}{2}} dx$ = $\frac{4}{15}$. Alternatively, you could do $\int_{0}^{1} y^{\frac{2}{3}}-y^{2} dx$ = $\frac{4}{15}$ $ HINT: Draw a sketch of both functions and determine their points of intersection, which will define your...

Ah, so what I did was assume that there was some sort of ordering of the coins within each of the boxes as well, which led me to believe they could be arranged in 10! ways, whereas we are only considering the ways in the coins can be assigned to the box (repitition allowed), hence 4^10 ways...

OK, then how would you do the following question (from Terry Lee 3U Ch7 Q24(b))? In how many ways can 9 different rings be arranged on the 4 fingers (excluding the thumb), assuming that no fingers can be empty? Answer says 8C3 X 9! = 20 321 280, where they have arranged the 9 different rings...

Wouldn't it still be 10! ways to arrange the coins if the boxes were still distinct?

It is because you will start to double count the combinations i.e. there will exist pairs of combinations which are exactly the same. So for instance, you may have P1, T1, P2, then somewhere else you will have P2, T1, P1 which is just a permutation of the first set (try some examples for...

Is it (1) AT LEAST one trigonometry question and STRICTLY ONE probability question or (2) AT LEAST one of each? If it is the first case, then you would do 4C2 X 5C1 = 6 X 5 = 30. If it was the second case, you would do (4C2 X 5C1) + (4C1 X 5C2) = 30 + 40 = 70.

$ If $\log2=x$ and $\log3=y$ then: \\ $\log12=\log4+\log3=2\log2+\log3=2x+y$

If the coins are different then we will have to find all arrangements of the coins too, which would be 10!. Thus, the total number of ways would then be 9C3*10! = 304819200.

Let me try to explain this a bit intuitively.. Since we have 10 identical coins, it doesn't matter how they are arranged as this will only give the same arrangement. So we only care about how they are distributed amongst the boxes. We can think of the coins being arranged in a row, and...

I think you forgot to divide your expression by 3!, since your separators are assumed to be identical, so the answer should just be $ ${9\choose 3}$ $=84 . Also, I got 286 number of ways you can arrange the coins such that there is NO restriction, so 504 which is double 286 wouldn't really...