Melody

YES it will always work!

I don't know how much algebra you have done but....

See if this makes sense

Let the middle number be x

So the one just below it is  x-1   and the one just above it is  x+1

The product of these two will be

$(x-1)(x+1) = x(x+1)-1(x+1)\\ (x-1)(x+1) = x^2+x-x-1\\ (x-1)(x+1) = x^2-1\\ (x-1)(x+1){\bf{+1}} {= x^2}\\$

I agree, patterns in maths are COOL !       

there are 5*4 =20 ways to chose the 2 freshmem for the end of the line.   

Now there are 7! ways to order rest.  =  5040 ways

But how many of those 7! ways do NOT have any 2 sophomores standing together.

the order would have to be  SFSFSFS  there are 4!*3! = 144ways

5040-144 = 4896

4896*20 = 97920 ways

Thanks Saphia, 

nect time just ask me, via private message, to open the old post and you can answer there. 

I have linked it to the original post.   

Answered here by Saphia1123

There is really only 5 sides to be set because once you have one you automatically have its opposite.

Also roations are the same

so I think it is just 

4! if all the colours are used  that is  24 ways

If only 4 colours are used then I think it is  5C4 * 4C1 * 4!/2     or    5C4 * 4C1 * 

1  =  5*4*12 = 240 ways

If only 3 colours are used and 3 are the same then maybe it is  5C3 * 3C1 * 4!/3! = 10 * 3 * 4 = 120ways

If 3 colours are used and 2 have 2 each then   i'm going for     5C3 * 3C2 *  4!/(2!2!) = 10*3*6 = 180 ways

If 2 colours are used then you have to look at each senario that is  1:4,  2:3  splits

1:4,   split      5C2*2 * 2C1 * 1 = 10 * 2 = 20 ways

2:3    split       5C2* 2   * 4C1 = 10* 2* 4 = 80 ways

If only 1 colour is used that there is only 1 possibility.   

24+240+120+80+20+80+1 = 565

It is highly likely that one or more of these are wrong.

Thanks Mathfreaker.    

I will point out however that It doesn't work if a and b are both 0. 

There needs to be a greater than OR equal to sign.

This question was auto blocked because you used the word 'suck.'

You need to be careful of your language online but  I have unblocked it for you. 

Is there a specific theorem to prove a^2 + b^2 > ab?

I will assume that a and be are in the set of real numbers.

If and a and b are both  0 then it is not true

If one is zero but not the other than LHS is positive and RHS is 0 so it will be true

If a or b (but not both) is negative then the RHS is positive and the LHS is negative so it will be true

So I am just left to prove it is true if a and b both have the same sign. (and not 0)

$(a-b)^2=a^2+b^2-2ab\\so\\ LHS=a^2+b^2\\ LHS=(a-b)^2+2ab\\ LHS=ab+[(a-b)^2+ab]\\ LHS=ab + \text{a positive number}\\ LHS >ab\\ LHS>RHS$

Think of the dice being different colours

There are 6*6 = 36 possible roll

How many ways can you roll a 3 ?

How many ways can you roll a 12?

Add those two together and the answer is that number over 36