Melody

Are 2,3 and 6 bases?

I think 2 3 and 6 are not bases but factors. But then i need to know what base the question is in.

I think you have left out key information.

Well, she could pull out 10 white then 1 black, 1brown 1 red, 1 blue.  And then the next draw would complete the second pair.

So I get 15.

Not sure but maybe ....  I admit that I really do not know what I am doing.

If you get the answer, with or without the logic, please share it here.

Take the Aces and hand them out 4! ways

Take the 2s and hand them out 4! ways

Take the 3s and hand them out 4! ways

4!*4!*4!

Or line all the cards up and hand them out in order  12! ways  I think I need to divide this by 3!3!3!3! = 6^4 = 1296

\(\frac{4!*4!*4!*1296}{12!}\\=\frac{4!*4!*1296}{5*6*7*8*9*10*11*12}\\ =\frac{24*24*1296}{5*6*7*8*9*10*11*12}\\ =\frac{3*2*1296}{5*6*7*9*10*11}\\ =\frac{*1296}{5*7*9*10*11}\\ =\frac{1296}{34650}\\ =\frac{72}{1925}\\ \approx 0.037\)

LaTex

\frac{4!*4!*4!*1296}{12!}\\=\frac{4!*4!*1296}{5*6*7*8*9*10*11*12}\\ =\frac{24*24*1296}{5*6*7*8*9*10*11*12}\\ =\frac{3*2*1296}{5*6*7*9*10*11}\\ =\frac{*1296}{5*7*9*10*11}\\ =\frac{1296}{34650}\\
=\frac{72}{1925}\\ 
\approx 0.037

This problem is not that hard to at least start.

You need to assume that nothing is painted other than the 2 sides that are painted blue.

Think about the box before you dismantle it.

How many blocks have 2 sides painted?

How many have only one side pained.

Maybe draw the cube or even just a 4 by 4 side to help you work it out.

Come back with your answers.

Please no one answer any more until Swimm makes a proper thought-through response.

Thanks guest,  this is a great contribution!

why not simplify the problem and let the 2 polynomials both be linear

m,n are both non zero and real.

\(f(x) = mx-4\\ g(x)= nx+k\\ \\~\\ h(x)= (mx-4)(nx+k) = ( some\; number)x^2 + ( some \;number )x   -4k\\ but\\ -4k=3\\ so\\ k=-\frac{3}{4}\)

So      g(0) = -3/4

Fine, do it then.  

Use elimination method

First add the 2 equations

\(x+x+xy^2+-xy^2=250y+-240y\\ 2x=10y\\ x=5y\\ \)

Now sub that into either original equation and solve for y there might be more than one answer.

See if you can go from there.

Don't forget to check your answers.

Your answer was answered yesterday bu EP  (possibly after you made this post but still about 24 hours ago)

AND you have not responded in any way.

Is it that you know how to complain but you do not know how to say thank you?

Yes thank you, I can see that your are correct, I have overcounted  

So I needed to looked at the 3 scenarios separately and then add the probablities together.

2 girls+5 boys

3 girls+ 4 boys

4 girls + 3 boy