Melody

I can only see 3 ways.

Say your have 2 reds and 4 blues.

1) You can have 4 blues together

2) You can have 3 blues together 

OR

3) you can have 2 lots of 2 blues together.

That is it I think.

Try proof reading you question.  I think there is a word missing.

Find the coefficient of u^2 v^9 in the expansion of (2u - 3v + u^2 - v^2)^9

[u(2+u)+v(-3-v) ]^9

$\binom{9}{n}\;[u(u+2)]^n\;[-v(v+3)]^{9-n}\\~\\ \binom{9}{n}\;[u^n(u+2)^n\;][-v^{9-n}(v+3)^{9-n}]\\~\\ (-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\ (-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\$

To get u^2   n must be 1

$(-1)^{9-1}\;\binom{9}{1}\;[u^1(u+2)^1\;][v^{9-1}(v+3)^{9-1}]\\~\\ =(-1)^{8}\;*9*\;[u(u+2)\;][v^{8}(v+3)^{8}]\\~\\ =9v^8\;[u(u+2)\;][(v+3)^{8}]\\~\\ $

The constant term for  (v+3)^8 = 3^8

So the u^2v^8 term will be in here

$9v^8\;[u(u+2)\;]3^8\\~\\ 9v^8\;[u^2+2u\;]3^8\\~\\ The \;\;u^2v^8 \text{ term will be}\\~\\ 9*3^8*u^2v^8 = 59049\;u^2v^8$

LaTex

\binom{9}{n}\;[u(u+2)]^n\;[-v(v+3)]^{9-n}\\~\\
\binom{9}{n}\;[u^n(u+2)^n\;][-v^{9-n}(v+3)^{9-n}]\\~\\
(-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\
(-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\

(-1)^{9-1}\;\binom{9}{1}\;[u^1(u+2)^1\;][v^{9-1}(v+3)^{9-1}]\\~\\
=(-1)^{8}\;*9*\;[u(u+2)\;][v^{8}(v+3)^{8}]\\~\\

9v^8\;[u(u+2)\;]3^8\\~\\
9v^8\;[u^2+2u\;]3^8\\~\\
The \;\;u^2v^8 \text{ term will be}\\~\\
9*3^8*u^2v^8 = 59049\;u^2v^8

100 000    to   999 999   is   900 000 numbers altogether

6 numbers that add to 8

The sum will be teween  8 and   54

so the sum could be   8,16,24,32,40,48,

800 000

710 000

620 000

611 000

530 000

521 000

511 100

440 000

432 000

431 100

422 000

421 100

411 110

332 000

331 100

322 100

321 110

311 111

222 200

222 110

221 111

There is an awflu lot to count. I guess there is a better method

970 000

961 000

952 000

951 100

943 000

942 100

941 110

933 100

932 110

931 111

922 210

922 111

...

-----------------------

The sum of  the 6 digits must be between 1 and 54  

6 of those are divisable by 8

If they were all equally likely outcomes then the answer would be  6/54. Trouble is they are not all equally likely outcomes.

there are 900 000 sums altogether.   Ifthey were equally likely it would be   6/54 * 900 000 = 100 000  but they are not equally likely.

Yea, I don't know.   

stars and bars

15C7 = 6435 possible outcomes

Give Dhruv her 2 chos,

now you have  6 friends and 7 chocolates

12C6 = 924 favourable outcomes

924/6435 =  28 / 195    =approx  14% chance

ok,

You need to be careful to proofread your questions. 

you have wasted your own and many other people's time.

If you still want an answer to your 'real' question I suggest you repost it properly on a new thread.

It is not an equation, it is a sum.

There is an infinite number of integer solutions.  Ths smallest is 2 and every integer larger than 2 is also a possible solution.

Try proofreading your questions Jason!

How many integers can be expressed in base 5 using exactly three digits

the digits are 0,1,2,3,4   repeats allowed but cannot start with 0

smallest = 100 base 5 = 5^2 base 10 = 25

biggest = (1000-1)base 5 = 5^3-1 base10 = 124 base 10

25<= [number base 10] <= 124

and expressed in base 7 using exactly two digits?

smallest = 10 base 7 = 7 base 10

biggest = 100-1  base 7= 7^2-1 base 10 = 48

7<= [number base 10] <= 48

The intersection is

25<= [number base 10] <= 48

48-25+1 = 44

Hi guest answerer.

Looking at in as different cases was a good idea

But

One of your early statements is 

"

Case 1: The first mailbox has 3 letters.

In this case, there are 3 letters left to be placed in the remaining 3 mailboxes."

there is only 4 letters, if three go in the first letter box then there is only 1 more letter.