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Re: HSC 2016 MX2 Combinatorics Marathon Another poker question: In how many ways can you be dealt a card-high hand (ie. not even a pair ..... a nothing hand)? (assuming of course that order is not important) Do this without counting all the other hands and subtracting from the total.

Re: HSC 2016 MX2 Combinatorics Marathon No takers? It really is quite easy once you look at it the right way. Definitely a 3U question.

Re: HSC 2016 MX2 Combinatorics Marathon In a particular maths test sat by a class of 31 students, no two students get the same mark. 11 of these students are randomly chosen. The medians of the class and the chosen group are calculated. What is the probability that the two medians are equal?

As in Christianity to Judaism.

Or more generally: \text{Since }z \overline z=\left|z \right|^2 \text{Then }\frac1{\overline z}=\frac z {\left|z\right|^2} \text{So }\text{Re}\left(\frac1{\overline z}\right)=\frac{\text{Re} (z)}{\left|z\right|^2} \text{From which the result easily follows}

Re: HSC 2016 MX2 Combinatorics Marathon It took me a while to figure out you had cross multiplied. I was thinking more of a probability story: LHS is the probability that if you choose k from the 2n, they will all come from a specified subset of n. RHS is the probability that if you...

Re: HSC 2016 MX2 Combinatorics Marathon A reminder that there are still two questions unanswered: \text{(1) (a) Show using algebra that: }\frac{\binom{n}{k}}{\binom{2n}{k}}=\frac{\binom{2n-k}{n}}{\binom{2n}{n}} \text{ (b) Explain this result in terms of a probability/counting...

Re: HSC 2016 MX2 Combinatorics Marathon Alas I did not. I found the 4 door version in a book, but decided to make it one step larger. Yes you are correct. This example appears early in the book, and they say that it will be solved later using the law of total probability. I am not sure how to...

Re: HSC 2016 MX2 Combinatorics Marathon Since nobody seems to want to answer my previous question, a spin on the Monty Hall problem: This game is played with 5 doors. After you make your initial choice the host opens a door that doesn't have the car, then gives you the option to switch to one...

Re: HSC 2016 MX2 Combinatorics Marathon Back to an Ext 2 level question: \text{(a) Show using algebra that: }\frac{\binom{n}{k}}{\binom{2n}{k}}=\frac{\binom{2n-k}{n}}{\binom{2n}{n}} \text{(b) Explain this result in terms of a probability/counting story}

Per-say?? Surely you mean per se.

Re: HSC 2016 MX2 Combinatorics Marathon Of course these lectures are not free. 35 lectures = 35 cans of bourbon.

Re: HSC 2016 MX2 Combinatorics Marathon Thanks for that. Looks like I have another series of YouTube lectures to view.

Re: HSC 2016 MX2 Combinatorics Marathon Thanks for the link. Apparently they are not derangements per se, but "partial derangements". It looks somewhat more complicated than a standard derangement (where ALL objects must be in the wrong position).

Re: HSC 2016 MX2 Combinatorics Marathon I have a question which I am having difficulty analysing. n balls are numbered 1, 2, 3, ..., n. The balls are placed randomly in a straight line, each ball sitting in a position 1, ..., n. Let P(n, k) be the probability that there are exactly k balls...

Still waiting for people to explain why the "evidence" they claim to see for a god when observing nature specifically points to their version of god and not a different one (including the Flying Spaghetti Monster).

How do you know that what you see is not evidence for the existence of the FSM? (Do your best to avoid circular reasoning this time.)

And if such an all-powerful being refuses to illustrate its power, or even its existence, how do I begin to believe in it? I believe .... I believe .... I believe .... I believe .... I believe .... I believe .... ah - there it is ...... now give me my 72 virgins.

That's right ..... a circular argument is required.

Still waiting for the photo showing god playing with those items.