Melody

You are welcome :)

You need to draw the table and put the people in follow what I am saying.

Put a particular child at the head of the table. that child represents the head of the table so roation has already been accounted for.

Now you can have an adult on each side of this child then the other child adults places are set.   

3! arrangements for adults, 2 arrangements for children = 3!*2 = 12

or

You can have an adult on one side and a child on the other side  then the other A/C are set

3! * 2 = 12

Total = ?

I think there are a couple of different ways to take lthis question.

Looking at the graph I can see that  b=a+2pi

From a to b is a whole wavelength of  cos a   and a half a wavelength of cos (a/2)

This means that

1-d = c - - 1

1-d= c+1

c=-d

\(cos\;a= \frac{1}{8}\\ cos(\frac{a}{2}+\frac{a}{2})=\frac{1}{8}\\ cos^2(\frac{a}{2})-sin^2(\frac{a}{2})=\frac{1}{8}\\ cos^2(\frac{a}{2})-(1-cos^2(\frac{a}{2}))=\frac{1}{8}\\ 2cos^2(\frac{a}{2})-1=\frac{1}{8}\\ 2cos^2(\frac{a}{2})=\frac{9}{8}\\ cos^2(\frac{a}{2})=\frac{9}{16}\\ cos(\frac{a}{2})=\pm\frac{3}{4}\\ \text{discount the positive answer}\\ c=cos(\frac{a}{2})=-\frac{3}{4}\\ \)

\(d=\frac{3}{4}\)

Hi YAD

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(note: I had to unblock this answer,  maybe because it was to long.  Not sure why, )

a^3 + 3ab^2 = 679 and a^3 - 3ab^2 = 673

\(a^3 + 3ab^2 = 679\qquad and \qquad a^3 - 3ab^2 = 673 \)

add them and get

\(2a^3= 1352\\ a^3=676\)

factor(676) = 2^2*13^2  not much help

subtract them

\(6ab^2=6\\ ab^2=1\\ \sqrt[3]{676}\;\;b^2 = 1\\ b^2=676^{-1/3}\\ b=676^{-1/6}\\ b=26^{-1/3}\\ \)

\(a-b=\sqrt[3]{676}-\sqrt[3]{\frac{1}{26} }\)

I could easily have made a careless mistake but I strongly think you may have made a copy error when you wrote the question in.

I would list out the possible number conbiations and then work out how many numbers I can make with each combination.

The maximum number of 3s is  3  that is  3331   there are 4 ways to order this

Now how many wars can there be two 3s

3,3,2,2                4!/(2!2!) = 6ways to order

3,3,1,1,2             5!/(2!2!)=30 ways to order

3,3,1,1,1,1,         6!/(2!4!) = 15 ways to order 

How many ways can there be one 3s

3,2,2,2,1             5!/(3!) = 

3,1,1,2,2,1           6!/(3!2!)=

3,1,1,1,1,2,1            7!/(5!)=

3,1,1,1,1,1,1,1         8!/7! =

Ok no more combinations use a 3

now go through a similar process with the 2s

starting with

2,2,2,2,2

etc

When you have finished add them all together.

Hint:

Use  probability contour mapping

Let the piece be    x, y-x and  6-y     units ong ... They add up to 6

b)

give them one each to start with then there are 15 stickets left

lie up the stickers  divide them into  5 bundles (some bundeles can be empty) the order of the bundles count becasue they are to be given to the five friends who are standing in a row.

 You need 4 dividers to do this

so you have 15+4 items = 19

19C4 = 3876 ways.

Use the same logic to get a)

This technique is called stars and bars method if you want to look that up.

No one can help you if you do not proofread your questions

Presenting nonsense helps no one.

12C3 = 220