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For the question An eight letter word is constructed from EEEEEEGG. What is the prob it begins and ends with an E? Method 1 Total number of eight letter words=8!/(2!6!)=28 Number that start and end with an E 6!/(2!4!)=15 Answer 15/28 Method 2 Prob(start E)xProb(end...

But we almost always answer these questions via counting!! An interesting point. If we ask the question What is the probability that an 8 letter word constructed from the 8 characters EEEEEEGG begins and ends with an E? then both methods agree! Why does the usual method of...

!!!!!!!!!!!!!!!!!!!!!! But the answer is NOT 4/11 !!!!!!!!!!!!!!!!!!!!!!! Imagine that I change the question so that the eight characters are replaced by 1000 E's and 2 G's. You will still get 4/11 with the above attack but the correct answer is clearly very close to 1. The problem...

Close!

Here's a strange one! a) A 4 letter word is to be constructed using the 8 characters EEEEEEGG. In how many ways can this be done? b) How many of the words in a) begin and end with an E? c) What is the probability that a 4 letter word constructed from the 8 characters EEEEEEGG begins...

Question is drawn from a 2004 selective school trial! Though there is no guarantee that the solution was correct. Either way it is a basic counting problem and hence in the syllabus

You would imagine that this (ie 77) is the correct answer but it is not!! Think about it!!

Here is a spooky little question for everyone preparing for the trials. In how many different ways can six identical black marbles and six idententical white marbles be arranged in a circle? Think carefully abvout your answer

A proof within the syllabus??

Anyone have a proof within the syllabus? What is the radius and centre of the new circle?

More precisely suppose that |z-a|=r (z lies on a circle of radius r centred at a) and |a| is not equal to r (circle does not contain 0) prove that the locus of 1/z is also a circle. Its trivial of course when a=0 I'm after a variety of different proofs (no insane determinant...

Prove that if z ranges over a fixed circle in the complex plane (not containing 0) then the locus of 1/z is also a circle

You need to show that a less than b implies a^3 less than b^3 Best not to use calculus

Prove that y=x^3 is an increasing function over the entire real line.

It's very interesting to see such varying approaches on this one. Anyone have any more proofs that are distinctly different?

Assuming usual naming around the circle angle 0 z1 z3=angle 0 z2 z3 (Why) Convert above to a statement re args show that arg(1/z1-1/z2)=arg(1/z2-1/z3)(mutatis mutandi)

But Gardner does not say a is not equal to b! He says they are equal to 4 decimal places and thus may be equal

I'm afraid it does! If a=b then a=b to 4 decimal places!

The first bold is just rewriting a=g-kv (tease out the g) for the second remember that the int of acc wrt to t is vel and the integral of vel wrt t is posn! Vg is just a constant so the int of Vg wrt t is Vgt.

Well I agree with that completely! It is totally consistent with what I've been saying.