Melody

Start by sorting them into bigger than 1 and smaller than 1.

then come back again if you need too.

thanks Builderboi.

That makes some sense.  

I suppose it could be an interesting assignement.

Something heavier than paper, maybe some reasonably think cardboard would give less error. 

Hi asinus,

That was good thinking.

But I graphed your equation using Desmos and there was a solution

Nice answer guest.    

Becasue the two that are not allowed to go together have been taken out 14-2 = 12

What Alan and I have said is

"Lets see how many of the 1001 combinations CONTAIN both those 2 people.  "

So now we have already chosen those two candidates.

We have 12 candidates left and we have to choose 2 of them  12C2

That is how many possible delegations of 4 that contain the 2 people who do NOT want to be together

That is the number of possible delegations that must be NOT counted

The number left AFTER these ones are NOT counted are removed      14C4  -  12C2

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

Lets's look at an easier questions,

We'll chose 2 letters from ABC        

3 letters 2 to chose that is 3C2=3 ways        AB, AC, BC

BUT      A abd B cannot both be chosen

How many of the 3 ways conatin both A and B       

there is one left to choose from and we need to chose 1 so there must be 1 combination with both A and B

(3-2)C(3-2) = 1C1 = 1

All the other combinations do NOT conatin both A and B

what is left is         3C2- 1C1 =  3-1  = 2

\displaystyle \int [ln(x)]^2 dx

\(\displaystyle \int [ln(x)]^2 dx\\ =\displaystyle \int [ln(x)] [ln(x)] dx\\ u=lnx\qquad v'=lnx\\ u'=\frac{1}{x}\qquad v=?\\~\\ ---------------------\\ \displaystyle \int [ln(x)] dx\\ =\displaystyle \int 1*[ln(x)] dx\\ p=lnx\qquad q'=1\\ p'=\frac{1}{x}\qquad q=x\\ so\\ \displaystyle \int 1*[ln(x)] dx\\ =xlnx-\int \frac{1}{x}*x\;dx\\ =xlnx-\int 1\;dx\\ =xlnx-x\\~\\ \text {So v=xlnx-x}\\~\\ ----------------\\ \)

Going back to the beginning

\(\displaystyle \int [ln(x)]^2 dx\\ =\displaystyle \int [ln(x)]  [ln(x)] dx\\ \qquad u=lnx\qquad v'=lnx\\ \qquad u'=\frac{1}{x}\qquad v=xlnx-x\\~\\ =uv-\int vu' dx \;\;+c \)

And you can finish it 

LaTex:

\displaystyle \int [ln(x)]^2 dx\\
=\displaystyle \int [ln(x)]  [ln(x)] dx\\
u=lnx\qquad v'=lnx\\
u'=\frac{1}{x}\qquad v=?\\~\\
---------------------\\
\displaystyle \int [ln(x)] dx\\
=\displaystyle \int 1*[ln(x)] dx\\
p=lnx\qquad q'=1\\
p'=\frac{1}{x}\qquad q=x\\
so\\
\displaystyle \int 1*[ln(x)] dx\\
=xlnx-\int \frac{1}{x}*x\;dx\\
=xlnx-\int 1\;dx\\
=xlnx-x\\~\\
\text {So v=xlnx-x}\\~\\
----------------\\

Hi Solviet,

You are still here.

the reason you are not on the leaderboard is becasue you have not posted anything for over a month.

If you make a post you will reappear   :)

Have you forgotten your password?

Lets see

a)  4 people  are chosen from 14       

   That is just 14 choose 4 = 14C4  = 1001     (it is not really choose, C stands from combinations)

   You say you get that?

b)   2 of the candiates do not want to go together.

        ok  Let's say those 2 are BOTH chosen

        Then there will be 2 chosen out of the remaining 12  which is    12C2  = 66

        BUT those are the combinations that are NOT allowed so the number of possible combinations will be

          14C4 - 12C2   = 1001 - 66 = 935

Does that help?

But the answes are already there, arn't they?

beats me how you can measure an AREA using a kitchen scale.