Melody

You have not said how many steps can be taken at a time.

I'll just look for a pattern

2 animals 2 ways

3 animals 3! ways

4 animals 

ABCD

b ac c can go anywhere   3*3*2*2 = 36 ways

ABCDE

If A goes to the middle then   BC (or CB) A DE(or ED)    4  

If A stays put 36 ways

If A goes second  

Now it is already getting all too much.

2. An elaborate code consists of left and right arrows which are placed together to represent words. There are spaces between words. For example, the code of three left and four right arrows arranged with two spaces: >> <> represents a three word code phrase. How many different three-word code phrases in this system use 8 letters?

I do not understand the explanation in question 2. I have no idea what is being described.

1. Three friends: Al, Bill, and Carl, are painting their chests with the word CAVALIERS for a basketball game, with at least two letters painted on each friend’s chest. For example, Al might get “CA”, Bill “VAL”, and Carl “IERS”, or Carl could get “CAVA” while Bill gets “LI” and Al gets “ERS”. How many ways can the three friends get their chests painted so that when they stand in the correct order, they spell out the word CAVALIERS?

CA   VELIE   RS 

Middle

VELIE

VELI

VEL

VE

ELIE

ELI

EL

LIE

LI

IE

So that is 10 ways  if the order the boys stand is is A, Bil Carl

If they can stand in any order then there are   10*3*2 = 60 ways.

Suppose that we have an equation y=ax^2+bx+c whose graph is a parabola with vertex (3,2), vertical axis of symmetry, and contains the point (1,0).  What is (a, b, c)?

roots are    \(x = {-b \pm \sqrt{b^2-4ac} \over 2a} \)

axis of symmetry is   

\(x= \frac{-b}{2a}\\ \frac{-b}{2a}=3\\ -b=6a\\ b=-6a\\ \)

\(y=ax^2+bx+c\\ y=ax^2-6ax+c\\ sub \;\;in \;\;(1,0)\\ 0=a-6a+c\\ 0=c-5a\\ c=5a \)       

\(y=ax^2-6ax+c\\ y=ax^2-6ax+5a\\ sub\;in\; (3,2)\\ 2=9a-18a+5a\\ 2=-4a\\ a=-\frac{1}{2}\\ \)

a=-0.5

c=-2.5

b=3

I do not have time to check so you had better do it :)

Thanks Alan,

I just fancied trying to show this graphically. :)

I used a variable k and it makes no difference :)

Here is a diagram that i draw.

It might be a bit hard to understand but i have drawn a quarter of the rectangle and the shaded bit is where the point will be closeset to the vertice than to the middle. It is pretty clear that exactly half the time it will be closer to a vertice and half the time it will be closer to the centre.

Lightning,

You are presenting high level questions so you are advance enough to learn how to present them properly.

I cannot be bothered interpreting this and I have read other people make similar comments.

The stuff between the dollars signs is in Latex.

Open the latex box in the ribbon, past the latex into there and press ok. Do not include the dollar signs.

Practice a little and you will find it easy.

If you need better instructions you can ask.

I think that this is what you mean.

\( csc\left[arctan\left(\frac{\sqrt{9-u^2}}{u}\right)\right]\\ \)

Draw a right angled triangle.   Mark an acute angle as theta.

\(tan\theta=\frac{\sqrt{9-u^2 } }{u}=\frac{opp}{adj}\\ hyp=\sqrt{9-u^2+u^2}=3\)

\(csc\theta = \frac{hyp}{opp}=\frac{3}{\sqrt{9-u^2}}\)

But u can be positive or negative     -3

so 

\( csc\left[arctan\left(\frac{\sqrt{9-u^2}}{u}\right)\right]=\pm\frac{3}{\sqrt{9-u^2}} \)

I think that is ok :)

Thanks Ginger   

I do not really understand your question, I think it is incomplete but perhaps you only want this part of it.

Given f(n)=1/(n+1)+1/(n+2)+...+1/(3n+1). When you want to prove the question using the mathematical induction method. For step 2, if n=k, you want to validate when n=k+1, then the relationship between f(k) and f(k+1) will be?

\(f(1)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\ f(n)=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}+ ....+\frac{1}{3n}+\frac{1}{3n+1}\\ f(k)=\frac{1}{k+1}+\frac{1}{k+2}+....+\frac{1}{2k}+ ....+\frac{1}{3k}+\frac{1}{3k+1}\\~\\ f(k+1)=\frac{1}{k+2}+.......\frac{1}{2k+2}+......\frac{1}{3k+3}+\frac{1}{3k+4}\\ f(k+1)=\frac{1}{k+2}+.....+\frac{1}{3k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\\ f(k+1)=\left[\frac{1}{k+2}+.....+\frac{1}{3k+1}\right]+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\\ f(k+1)=\left[f(k)-\frac{1}{k+1}\right]+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\\ f(k+1)=f(k)+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}-\frac{1}{k+1}\\ \)

check:

\(f(1)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\ f(2)=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\\ f(2)=f(1)+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{2}\)

Well that looks promising.  :)