Alan

4x²+40x+100 = 4(x² + 10x +25) = 4(x+5)², so for this to be zero x must be -5.

Ok CreeperCosmo, except that you should have used the word "volume" where you have "area".

It's also worth noting that, in general, you need to consider the shape of the individual blocks as well as their volume.  For example, suppose each block was 12" by 1" by (1/2)".  The individual volume is still 6 cubic inches, so the volume ratio would still suggest you could get 36 of them in the box.  However, because of their shape you wouldn't even get one in the box.  

The volume ratio tells you the maximum conceivable number that can go in the box, but you need to take into account the shape of the individual pieces to decide how many will actually fit.  

In the case of the original question here, the shapes stack nicely to fill the box completely.

Hmm! The notation used in the question suggests that the first term is a=6 and the geometric progression ratio is r=-3.

If this is the case then the sequence goes

a = 6

ar = -18

ar² = 54

ar³ = -162

GoldenLeaf's approach was almost right, but he should have had 5, 4, 3, 2 and 1 in the numerators.

There are 5 possible choices for the first number, so probability of this is 5/40.  For each of these

there are 4 possible choices for the 2nd number, so probability of both 1st and 2nd is (5/40)*(4/39).  For each of these ... etc. to get the overall probability as

$${\mathtt{p}} = \left({\frac{{\mathtt{5}}}{{\mathtt{40}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{39}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{38}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{37}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{36}}}}\right) \Rightarrow {\mathtt{p}} = {\mathtt{0.000\: \!001\: \!519\: \!738\: \!361\: \!8}}$$

or p≈1.52*10^-6  

1/p = 658008

3 parts are grey and 1 part is black.  This means there are 4 parts in total.  Therefore the black fraction is 1/4. Multiply this by 100 to get the percentage: 100*1/4 = 25%

I guess you are referring to the calculator on this site, which, when calculating factorials, has a maximum of 170! for which it gives a result (above that it just returns infinity).  The word "only" seems inappropriate - there are over 300 digits in 170! - how many do you want?!!!

Melody said: Can someone please remind me how to put the last 2 answers into the Input=Result.

npr(100,2)  gives 

$${\left({\frac{{\mathtt{100}}{!}}{({\mathtt{100}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{9\,900}}$$

2*ncr(100,2) gives

$${\mathtt{2}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{100}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{100}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{9\,900}}$$

I misread the question!

for 10*20-(20/10)+21
if I do braket first, (20/10)=2
then multiplication comes, so, 10*20=200

You are right to here, but your next step is wrong.  You should have 

200-2+21 = 200 + 19 = 219 because -2 + 21 is +19 not -23.