Melody

An oldie but a goodie.

Thanks Chris,

And Thanks Heureka for bringing me back here.   

I never answer questions when they are presented with unrendered LaTex.

So the question is:

The line with equation -a+3b=0       (is this one meant to be zero ?) 

conincides with the terminal side of an angle Ø

in standard position and sin Ø   < 0

What would the value of cos Ø  ?  If sin is negative then cos is positive.  (4th quadrant)

It still makes no sense to me.

Does this comewith a picture?

lines usually use x and y.

I can explain more but I want to stress that I am not sure if this is completely correct. 

My answer is the same as CPhill got with his formula but I am still not convinced that my method is right.

It is the things you are questioning that i am not sure about  but here is your explanation.

(b) I also bought 7 different postcards. How many ways can I send the postcards to my 3 friends, so that each friend gets at least one postcard?

The original answer was 15, that was if all the cards were the same.

possible numbers of bags each (total 7)   

511     3  ways if all the same   7*6=42              3*42=126

the first person can get any of the 7 postcards, the second person gets any of the 6 remaining, the 3rd person gets the rest.

Hence 7*6* the number of ways if the cards had been all the same.

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421     6  ways if all the same   7*6C2=105       6*105=630

the first person can get any of the 7 postcards, the second person gets any 2 of the 6 remaining, the 3rd person gets the rest.

Hence 7*6C2 times the number of ways if the cards had been all the same.

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331     3  ways if all the same   7*6C3=140       3*140=420

The first person can get any of the 7 postcards, the second person gets any 3 of the 6 remaining, the 3rd person gets the rest.

Hence 7*6C3 times the number of ways if the cards had been all the same.

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322     3 ways if all the same   7C2*5C2=210     3*210=630

The first person can get any 2 of the 7 postcards, the second person gets any 2 of the 5 remaining, the 3rd person gets the rest.

Hence 7C2*5C2 = 210 times the number of ways if the cards had been all the same.

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Total   15 ways if all the same 1806 ways if all are different

nCr is a little formula, (which most times you do not even have to know, the formula I mean)

Anyway

nCr  which is available on just about any calculator means

how many ways can r objects be chosen from n objects.

So if you have 10 people and you want to know how many different groups of three you can make that would be 10C3 ways.

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Now look at the first question i did.

(a) While travelling abroad, I bought 7 identical bags of candy for 3 friends. How many ways can I distribute the candy to my 3 friends, so that each friend gets at least one bag of candy?

7 identical bags shared between 3 friends but they all get at least one each.

So really it is how many ways can 4 bags be given to three different people.

Here are the 4 bags      ****

I need to break them up into 3 groups. I will use bars to seperate them. I need 2 bars.

Person 1 gets the bags in front of the first bar

Person 3 gets the bags behind the last bar

Person 2 gets the bags in the middle.

eg

||****   

This would be one possibility.  Person 1 and 2 get none and person 3 gets all four of the bags.

The bars can go anywhere.

So  the number of possibilities is     6 items, choose 2.  Those two will be the bars, (people)

6C2 ways.

this is called the stars and bars method by the way.

Thanks for your responses :)

Ok thanks EP. It keeps saying my security for thsi site is not secure and wants me to log in on its own boxes.

I close it and it lets me keep going but then it rears its ugly head a few clicks later. 

It is annoying.

Jenny I think that sometimes happens when your picture is copyrighted,

Can you think of a different way to put it in. Like maybe from a screenshot?

Or Just give the link.