Melody

If you know the textbook/teacher answer can you please share it with us.

In how many ways can I place 5 different beads on a bracelet if rotations and flips of the bracelet are not treated as different arrangements?

Without thinking to much I think it would be    4!/2 = 12 ways.

Thanks Sudsw12,

I will examine what you have said in detail when I have a little more time.

Thanks for answering and  doing this research for me.

It is much appreciated.

Do you understand what Rom wants you to do?

graph each side seperately, put y= out the front of each side and see where the graphs cross.

The x value of intersections will be your solutions.

Good thinking :)

If f(x) is a polynomial of degree 7, and g(x) is a polynomial of degree 7, then what is the product of the minimum and the maximum possible degrees of f(x) + g(x)?

if f(x)  = - g(x)    then  f(x)+g(x)=0    degree zero.

Most times though the f(x)+g(x) would have degree 7

7*0=0 

Oh, and if you have an answer provided please explain that, and give it, in your original question.

You should not give us less information than you have for yourself.

You should also explain what you have already done in order to attempt to answer the question yourself.

ALSO

You have asked far too many questions in a short space of time. 

This is an abuse of the forum.

You need to ask a question and learn from the answer before you ask the next question.

You also need to interact with answerers if you want to make any degree of a positive name for yourself here.

You need to look at thre different cases.

1) The pairs all sit infront/behind.

See how many ways the first row can be filled and the second row will automatically be accounted for.

I think that is just 10*8*6*4*2 = 3840 possibilities.

2) 1 pair sit in each row side by side the others are infront/behind

Side by side couples:

there is 4 possible places to sit the pairs.

There are 10 ways to chose the first person of the first pair and then 8 ways to chose the first person of the second pair

So I guess that is 8*10 *4= 320 Ways.

Now there is 3 paired people/places left  6*4*2=48 ways to sit them

320*48=  15360 possibilities

3) 2 pairs sit in each row, the others are infront/behind.  

The individual one can be in the 1st 3rd or 5th positions. 

There are 5 ways to choose the odd pair so that make it 3*5*2= 30ways to chose.

Now there is 4 couples to sit side by side. and there are 4 paired places to put them.

4!*2^4=384 

30*384 =11520  possibilities

3840+15360+11520 = 30720 possibilities.

Since this answer agrees with your book it has a fair chance of being correct.

Thanks Heueka and Sudsw12

I was looking for a shortcut but I didn't quite get there :)

HOW do you know this

Any positive integer n is congruent, modulo 9, to the sum of its digits.  I have not seen that before.

I know that if the sum of the digits is 0 mod 9 then n will also be 0 mod 9.   I think that is reasonably common knowledge.

But

To I have not seen that extend to any sum of digits before. 

I wonder how difficult that is to prove.  I have not tried yet.