Melody

That is impressive, but your 30 still does not make sense to me.  

Would you like to try and explain your logic there?

EDIT

Thanks for your input guest. I have worked it out using my own logic now.

30 makes perfect sense.

 fair 6-sided dice are rolled. What is the probability of all 6 numbers showing (1, 2, 3, 4, 5, 6)? The remaining 3 numbers can take any value. Thank you.

Here is an adaption of my first answer.

I have corrected all silly errors and I have corrected one logic error.

Now it is correct

Number of permutations in total = 6^9 = 10, 077,696

3 doubles the others different =        \( 6C3 *\frac{9!}{2!2!2!}=45360=20*45360 = 907200\)

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

The next one was the one that gave me the most trouble but it should not have

the 6C2 is because to of the numbers must be chosen as the duplicate ones.

Then i had to double it because one occured 3 times and the other only 2 and it could be either way around.

so

 1 double, 1 triple =       \(  6C2*2*\frac{9!}{3!2!}=30*30240=907200\)

-------------------------
1 quad = \(6*1*1*\frac{9!}{4!}=6*15120=90720 \)

907200+907200+90720 = 1905120 

1905120/10077696 = 0.1890432098765432

Approx   18.90%

Finally our answers agree and they also agree with Wolfram Alpha!

Micacle of miracles.

If every box must have a ball then it would be 4! = 24 ways

Assuming that some boxes do not have to have any balls then I think the answer is 

7C3=35 ways

Mmm

You might be right, it does sound logical,

BUT

if it is meant to be 6C3 (as you did)  then 30 (which we both got) is also wrong, it should be 6C2=15

Hi guest,

I made a number of careless errors in my first attempt.

I have now fixed the ones that I found.

Our techniques are basically the same.

But we have different answers for 3 doubles and 3 singles. Could you just check yours please?

Thanks guest for the attempt.  

I have not come up with a likely answer but I am reasonably confident that yours is not correct.

Please outline your working guest.