Melody

nCr 

$$^5C_2$$   would be entered into the web2 calc like this    nCr(5,2) 

It means, how many ways can 2 things be chosen from 5 if order does not count

(the C stands for Combinations).  

I'll enter it now,  it adds the maths but you don't need to worry about that.

$${\left({\frac{{\mathtt{5}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{10}}$$          See, there are 10 ways

Alan answered a much more complicated question.

He was using the numbers 1,2,3,4,5,6,7,8, 9, and 10

There are 10 ways to chose just one number  

That is 10C1 and you would put it into the web2 calc as  nCr(10,1)

There are 10C2 ways to choose 2 numbers  that is nCr(10,2)

...

there are 10C10 ways to chose 10 numbers (this will be = 1)  but it can go into the calc as  nCr(10,10)

Alan used this symbol     $$\sum$$     it means "sum of"

so his answer

$$\sum_{k=1}^{10}nCr(10,k)=1023$$

is the same as     10C1+10C2+10C3+10C4+10C5+10C6+10C7+10C8+10C9+10C10

Well Eloise you had better know how many ways you can house your elephants then!!      LOL

Hi Chris :))

Yes, you could learn this stuff MathGod it is not that hard

but I want you to concentrate mainly on arithmetic for now.   

If you study maths for long enough you will learn this stuff too Eloise :))

Hi again MathsGod,

Basically a log is a power.

For instance

    $$\\2^3=8\\\\
so\\\\
log_28=3$$

See how the answer to the log is the power   (or exponent) 

BUT

Your brother is correct, they are difficult to understand and you are nowhere near ready to deal with them yet!       

You are very welcome Lunar   

I could make a counting question out of this!

There are 10 elephants that I will class as identical, and 10 different homes that they can go to.

Any number of elphants can go to any of the homes.  How many ways can this be done?

The bars represent the homes and the * represent the elephants.  

The elephants are entering the door of the hous from the left side.  

This would  be one combination

*|***|||**||**|**|||

1 elephant in house 1

3 elephants in house 2

NO elephants in house 3 or 4

2 elephants in house 5

No elephant in house 6

2 elephants in house 7

2 elephants in house 8

No elephants in house 9 or 10

Now if I put the stars in different spots the the number of elephants in one or mor houses will be different!

The last bar has to be at the end so its position cannot change and it is not used in the calculation.

Altogether there are 19 stars and bars left and I want to know how many ways I can choose 10 of those to be elephants (stars)

That is 19C10   

$${\left({\frac{{\mathtt{19}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{19}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)} = {\mathtt{92\,378}}$$

I think that is right.  There are a lot of different ways our elephants can be housed!!

NOTE:  I could look at this a little differently

Altogether there are 19 stars and bars left and I want to know how many ways I can choose 9 of those to be houses (bars)

This would be    19C9

$${\left({\frac{{\mathtt{19}}{!}}{{\mathtt{9}}{!}{\mathtt{\,\times\,}}({\mathtt{19}}{\mathtt{\,-\,}}{\mathtt{9}}){!}}}\right)} = {\mathtt{92\,378}}$$

See, the answer is the same!!

Hi MathGod,

You and Alan have answered very different questions.

Alan has answered a combination question and you have answered a permutation question.

 I will give you a much more simple question to try and show you the difference

1a) How many Combinations are there of 2,3,4,5,6,7,8,9  All numbers must be used just once.

The answer to this is 1  

This is because order does not matter so there is only one way to choose 8 numbers from 8 numbers!

1b) How many Permutations are there of 2,3,4,5,6,7,8,9  All numbers must be used just once.

This is the question you answered and your answer of 8! was correct for this question.

With permutations the order matters that is why there are so many answers.

2A    How many combinations of 2 digits can you get from  4,5,6

Well lets see, there is   4,5   and   4,6   and  5,6   That is it!  The answer is 3 combination

2B   How many permutations of 2 digits can you get from  4,5,6

Mmm,      4,5    and   5,4     they are different for permutations because order counts!

there is also     4,6   and   6,4    and   5,6  and     6,5      and that is it.  SO there are   6 permuations