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We have two tasks here....choose the boxes....choose the b***s to go into those boxes...

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4 2 0

We have 3  ways to choose the first box......and we want to put 4 of the 6 b***s into it

3C(6,4)  = 45

Then, out of the remaning 2 boxes, we want to choose one of the two to put the remaining two b***s into

So 45 x 2 = 90

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4 1 1

Agian we have 3 boxes to choose which 4 b***s out of 6 we put into it

So 3C(6,4)= 45 ways

And we have two ways to select the next box and two ways to select which ball goes into that box...

The last box and ball are defaulted

So we have

2C(2.,1) = 4 ways

So 45 x 4 = 180

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3 3 0

3 ways to choose the first box......and we want to choose 3 of 6 b***s to put into it

3C(6,3) = 60 ways

Then, we have 2 ways to choose the next box to place the remaining three b***s into

So 2 x 60  = 120 ways

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3 2 1

3 ways to choose the first box....choose 3 of 6 b***s to put into it   = 60 ways  

2 ways to select the next box and we want to choose 2 of the 3 remaining b***s to put into it

2C(3.2) = 6

Again, the last box and ball are defaulted

So   60 x 6 = 360 ways

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2 2 2

3 ways to select the first box .... choose 2 of 6 b***s to go into it  = 3C(6,2) = 45

2 ways to pick the next box and choose 2 of the remaining 4 b***s to put into it

2C(4,2)  = 12

The last box and remaig two b***s are defaulted

So 45 x 12 = 540 ways

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Anyway, that's what I get......!!!

Hi GoldenLeaf,

Thanks chris :)

6 DIFFERENT B***S AND 3 DIFFERENT BOXES

THIS HAS BEEN EDITED

I BELIEVE CPHILL AND I ARE FINALLY IN TOTAL AGREEMENT

DIFFERENT BOXES - 3 WHITE AND 3 BLACK B***S.

0,0,6             3*1=3  choice

0,1,5              3!* the 1 could but black or white   6*2=12 choices

0,2,4              3!*the 2 could be BB, WW or BW     6*3 = 18 choices

0,3,3              3* the 3 could be  BBB, BBW, BWW or WWW but 2 of these are the same so 2 choices

                      (BBB,WWW)  or   (BBW)(WWB)     3*2= 6 choices

1,1,4              3* the ones can be   BB, WW or BW       3*3 = 9 choices

1,2,3              3!*    

                      If the first box has a B then the second can have BB, BW or WW

                      If the first box has a W then the second can have BB, BW or WW     6*6= 18 choices

2,2,2              1* If one box has WW then the other boxes must be BB and BW.

                     or    BW,BW, BW                                                                   1*2=   2 choices

$${\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{18}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{18}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{68}}$$

There is no guarantees that this is correct.

I might have done it slightly differently, but I believe your answers are correct, Melody....

Here, the boxes don't matter...one is as good as the other...!!!   The only thing  that matters is choosing which b***s to group together in one (or more) boxes. 

Another 3 points from me....!!!

Just I have some questions 

First of all if there is no point charge inside the cavity of the sphere the inner radius will carry any charge! 

Well, I got 7 ways if the boxes are all  the same so I am going to start with those solutions. (I hope I got them all. I think i did)

6 DIFFERENT B***S - 3 IDENTICAL BOXES

0,0,6       1 way

0,1,5        6C1 = 6 ways

0,2,4        6C2 = 15 ways

0,3,3       6C3 = 20 ways

1,1,4        6*5=30 ways

1,2,3        6*5C2 = 60 ways

2,2,2       6C2*4C2 = 15*6=90 ways

$${\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}{\mathtt{\,\small\textbf+\,}}{\mathtt{20}}{\mathtt{\,\small\textbf+\,}}{\mathtt{30}}{\mathtt{\,\small\textbf+\,}}{\mathtt{60}}{\mathtt{\,\small\textbf+\,}}{\mathtt{90}} = {\mathtt{222}}$$      

That is what I think but it may not be correct.

Thanks Chris

$$\frac{d}{dx}x^{4/3}=\frac{4}{3}\times x^{1/3}=\frac{4\sqrt[3]{x}}{3}$$

Yep....that answer seems reasonable, Melody......

3 points from me......

If the b***s are all the same and the boxes are all the same then the only difference is how many b***s are in the boxes.

I counted them and I get 7 ways

0,0,6

0,1,5

0,2,4

0,3,3

1,1,4

1,2,3

2,2,2

Since all the b***s and all the boxes are the same, the only difference can be is how many b***s  are in the boxes.

Yeah...check my math, Melody....you're not the ONLY one wearing glasses...!!! LOL!!!

YES the question does clearly state that - next time I really must wear my glasses when i am reading questions.

I probably misread this same word in my other answers as well which would make them all wrong.  

The joke is on me     ROF LOL

Thanks Chris. 

CPhill's second answer may well be correct - that is certainly how I should have started it but I have not worked it out all through to the end. :)

What the H**L???

How was I supposed to know WHAT you wanted???

I certainly didn't see the word "centroid" anywhere in your question...!!!

I thought you were wanting the area between the curves {silly assumption on my part...}

If you want a specific thing......maybe you should INCLUDE that specific thing when you post.....!!!!

Melody has assumed the b***s and the boxes are both indistinguishable....

But.....the question clearly states that they are both distinguishable....capable of being specificallly identified....

Thus. all the b***s are different and all the boxes are different....

(???)

t is common to both terms so you can write: t(9.75t + 1.8) = 0

This mean that either t = 0 or t = -1.8/9.75  

.

They are functions, each of which is the inverse of the other.  By definition ln(x) is log to the base e of x, and exp(x) is e to the power of x (where e is the number 2.718...).

They are functions, not numbers; though when x is a specific number they each return a number.

For example, suppose x is 2, then exp(2) =  7.389056098930650227230427460575...

Now suppose x is 7.389056098930650227230427460575..., then ln(7.389056098930650227230427460575...) = 2

ln reverses the effect of exp.

Suppose x is 5, then ln(5) = 1.6094379124341003746007593332262...

Now supoose x is 1.6094379124341003746007593332262..., then exp(1.6094379124341003746007593332262...) = 5

exp reverses the effect of ln.

.

There are 12 parts in the $72

Mahesh gets (7/12)  = $42

Jayrag gets (5/12) = $30

Imagine you are at a value of r that is much greater than R₂.  All you will be able to see is a point that has total charge q + Q.  

Well now, that is interesting.     LOL

I am not laughing at Chris, my answer could be wrong.  The 2 extreme answers are funny though.  :)

I got the same.  Miracles to happen sometimes.  LOL

My first answer was c**p....I'll try again

If the b***s and boxes are distinguishable

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We can put all 6 b***s into any box.....there are 3 ways to do this....put them into the first box, the second box or the third box

So 3 ways in this arrangement

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We can put 5 b***s into one box, 1 in another and 0 into the last

 But since the b***s are distinguishable, we can pick any 5 of the 6 to put into a box and we have 3 ways to choose that box. =  3C(6,5) = 18 ways

And we have 1 way to choose the next ball and 2 ways to select the next box = 2 ways

The last box is empty and determined by default

So 18 x 2 = 36 ways in this arrangement

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And we can put 4 b***s into one box and 1 in each of the remaining two.

So we have 3 ways to choose the first box and we want to select any 4 of the 6 b***s to put into that box. So 3C(6,4) = 45 ways

And we have 2 ways to choose the next box and 2 ways to choose the next ball = 4 ways

And the final box and ball are determined by default

So 45 x 2 = 180 ways in this arrangement

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And we can select 3 b***s in one box, 2 into another and 1 in the last

We have 3 ways to select the first box and we want to choose 3 of the 6 b***s to put into that box......so we have 3C(6,3)  = 60 ways to do that

Then, we have 2 ways to choose the next box to put 2 of the remaining 3 b***s into....so this is  2C(3,2)  = 6 ways

Again the last ball and and box are determined by default

So the total ways here are  60 x 6 = 360 ways in this arrangement

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And we can put 4 b***s into one box and two into another

We have 3 ways to choose the first box and we want to choose any 4 of the 6 b***s to put into this = 3C(6,4) = 45 ways

And we want to put the other two b***s into either of the remaining two boxes = 2 ways to do this

The last box is empty by default

So 45 x 2 = 90 ways in this arrangement

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And we can put 3 b***s into one box and 3 into another

There are 3 ways to choose the first box and we want to choose 3 of the 6 b***s

So 3C(6,3) = 60 ways

And there are 2 ways to select the next box to put the remaining 3 b***s into = 2

The last box is empty by default

So 60 x 2 = 120 ways in this arrangement

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And finally, we can put 2 b***s into every box

We have 3 ways to choose the frst box and we want to choose 2 of the 6 b***s to put into that box  3C(6,2)  = 45 ways

And then we have 2 ways to select the next box and we want to choose 2 of the remaining 4 b***s to put into this  = 2C(4,2) = 12 ways

The last 2 b***s and box remain by default.

So 45 x 12 = 540 ways

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So, to recap

6 0 0 =  3 ways

5 1 0 = 36 ways

4 2 0 = 90 ways

4 1 1 = 180 ways

3 3 0 = 120 ways

3 2 1 = 360 ways

2 2 2 = 540 ways

So   3 + 36 + 90 + 180 + 120 + 360 + 540 = 1329 ways

If it's the coordinates of the centroid of the area between the two curves that you want, then see the following:

.

no definitely not.