Melody

Thanks,

I should of thought of that but I didn't.   

Thanks for showing me     

Yes that is right EP. 

At least I agree with you and 2 great minds can't both be wrong ... or can they LOL.

How many ways are there to divide a group of 6 friends among the basketball team, the soccer team, and the track team? (Each team could have anywhere from 0 to 6 of the friends on it. Assume the friends are distinguishable.)

6 people into 3 groups.

6 0 0       3 * 6C6              = 3*1               =      3 ways

5 1 0       3! * 6C5           = 6*6                  =   36 ways

4 2 0       3! * 6C4            =6*15                =   60 ways

4 1 1       3* 6C4   *2       =3*15*2              =  90 ways

3 3 0       3* 6C3               =3*20              =   60   ways

3 2 1       3!* 6C3  *3C2    =6*20*3           = 360   ways

2 2 2       1*  6C2  *4C2    =15*6              =   90   ways

3+36+60+90+60+360+90 = 699

I wont guarentee that it is correct though.  

Hi Geno, let's see if i get the same :)

How many subsets of the set {1,2,...,11} have median 6?

1 2 3 4 5    6    7 8 9 10 11

The 6 is included

1+(5C4)^2+(5C3)^2+(5C2)^2+(5C1)^1+1

= 1+5^2    +10^2      +10^2    + 5^2     +1

= 1+25+100+100+25+1

=252

6 not included but 5 and 7 both included

(4C4)^2+(4C3)^2+(4C2)^2+(4C1)^2+4C0)^2

=1         +4^2          +6^2      +4^2     +1

=1+16+36+16+1

=70

5,6, and 7 not included but 4 and 8 inlcuded

(3C3)^2+(3C2)^3+(3C1)^2+(3C0)^2

=1+9+9+1

=20

4,5,6,7,8 not included but 3 and 9 included

(2C2)^2+(2C1)^2+(2C0)^2 

=1+4+1

=6

3,4,5,6,7,8,9 not included but  2 and 10 included

1+1

= 2

2,3,4,5,6,7,8,9,10 not included but 1 and 11 included

=1

252+70+20+6+2+1 = 351

idk, I never finished it.

$S_n=\frac{a(1-r^n)}{1-r}\\ S_{11}=\frac{1(1-0.2^{11})}{0.8}+\frac{-0.5(1-0.2^{11})}{0.8}\\ S_{11}=\frac{0.5(1-0.2^{11})}{0.8}\\ S_{11}=0.624999...\\ S_{11}\approx 0.62 \qquad \text{to 2 dec places}$

$1 - \frac{1}{2} + \frac{1}{5} - \frac{1}{10} + \frac{1}{25} - \frac{1}{50} + \frac{1}{125} - \frac{1}{250} + \cdots + \frac{1}{5^{10}} - \frac{1}{2 \cdot 5^{10}}\\ =1 + \frac{1}{5} + \frac{1}{25} +\frac{1}{125} + \cdots + \frac{1}{5^{10}} \\ - \frac{1}{2} - \frac{1}{10} - \frac{1}{50} - \frac{1}{250} + \cdots - \frac{1}{2 \cdot 5^{10}}$

the first line is the sum of a GP    a=1,          r=1/5       11 terms

the second is the sum of a GP     a=-1/2,      r=1/5      11 terms        (that dot is a times sign not a decimal point) 

Work them out and add them together.      

Show us your attempt

You make it look easy Alan      

Here is the answer you have provided.

It lost me after the first sentence ...   

(2003^2003+1):(2003^2004+1)=?

$(2003^{2003}+1):(2003^{2004}+1)\\ (2003^{2003}+1):(2003*2003^{2003}+1)\\$

Not much you can do with this. 

Should there have been more brackets in your question?