Compute the minimum number of seconds necessary for Kevin to evaluate the expression together by performing eight such multiplications 
For the NEMO, Kevin needs to compute the product
  $$9 \times 99 \times 999 \times ··· \times 999999999.$$
  Kevin takes exactly $ab$ seconds to multiply an $a$-digit integer by a $b$-digit integer. Compute the minimum number of seconds necessary for Kevin to evaluate the expression together by performing eight such multiplications.

It seems that there would be a lot of combinations and I can't seem to find a systematic way of finding the minimum.
 A: If we have any three integers $a,b,c$ with $n_1,n_2,n_3$ digits respectively and assuming that multiplications don't underflow in the sense that $ab,bc,ac$ has $n_1+n_2$, $n_2+n_3$ and $n_1+n_3$ digits then the time taken to multiply $(ab)c$ is
$$(n_1n_2) + (n_1+n_2)n_3 = n_1n_2+n_1n_3 + n_2n_3$$
which is symmetric in $n_1,n_2,n_3$ so the order we perform the multiplications in does not matter. As long as the assumption above continues to hold as we multiply more and more numbers in it follows that the order we do this in is irrelevent for any number of elements in our list. This is indeed the case here since (see the end for more details)
$$\frac{1}{10} < 0.9\cdot 0.99 \cdots 0.999999999  < 1$$
The total time is therefore the same as the time used to perform the multiplications in accending order which is
$$\sum_{n=2}^9 [9_n][9_19_2\ldots 9_{n-1}] = \sum_{n=2}^9 n\cdot [1+2+3+\ldots + n-1] = \sum_{n=2}^9\frac{n^2(n-1)}{2} = 870$$

${\bf More~details}$: We can always write an integer $a$ on the form 
$$a = a_{\rm frac} 10^{a_{\rm digits}}$$ where $a_{\rm frac} \in \left[\frac{1}{10},1\right)$ and $a_{\rm digits}$ is an integer (which is also the number of digits of $a$). For example $99 = 0.99\cdot 10^{2}$ so $99_{\rm frac} = 0.99$ and $99_{\rm digits} = 2$. Now if we have another integer $b = b_{\rm frac} 10^{b_{\rm digits}}$ then
$$ab = a_{\rm frac}b_{\rm frac}10^{a_{\rm digits}+b_{\rm digits}}$$ 
so as long as the product $a_{\rm frac}b_{\rm frac}$ is also in $\left[\frac{1}{10},1\right)$ we see that $ab$ will have $a_{\rm digits}+b_{\rm digits}$ digits. From this we see that $9\cdot 99 \cdots 999999999$ will have $1+2+\ldots + 9$ digits iff
$$\frac{1}{10} \leq 0.9\cdot 0.99 \cdots 0.999999999 < 1$$
which is indeed the case.
A: To the other answers that already give the correct answer, I would like to add a more thorough argument why the order of multiplications does not matter. Multiplication is not only commutative but also associative, which means that apart from permuting the terms one can also group terms together in different ways; one might imagine for instance that the depth of the evaluation tree has influence on the result, and this concern cannot be addressed by just comparing different ways to combine three terms.
Like Winther explains, the first thing to do is check that any product of some of these numbers, each divided by the next power of $10$ so as to be just less than$~1$, always gives a number between $0.1$ and $1$, from which it follows that whatever order of multiplication one uses, each multiplication of an $a$-digit number by a $b$-digit number will result in an $(a+b)$-digit number.
The following model then provides an equivalent problem, in which the accounting of the number of multiplications required is such that the independence of the grouping and order becomes clear.
Imagine some people initially partitioned into several groups. Assume that within each group people are all acquainted with each other. Two groups may meet, at which point all people from the first group shake hands with all people of the second group; then the groups unite to a new group in which again all people are acquainted with each other. If these were groups of $a$ and $b$ people respectively, then the process involved $ab$ handshakes and results in a group of size $a+b$; thus this exactly models the multiplication process in the question.
After repeating this process several times until there is only one big group, any pair of people that originally were in different groups will have shaken hands once, and this accounts for all the handshakes. Then it is clear that the number of handshakes does not depend on the precise manner of grouping that was applied to unite all the groups together.
Now it remains to count the number of pairs among those $1+2+3+\cdots+9=\binom{10}2=45$ initial digits that occur in different numbers. There are $\binom{45}2=990$ pairs in all, from which one must subtract the pairs within one of the numbers. The number to subtract is $\binom12+\binom22+\binom32+\cdots+\binom92=\binom{10}3=120$, and so the total number of digit-multiplications used in any way to multiply these numbers is $990-120=870$. 
A: Well it turns out that any path you take is the same. It can be said, that since we are doing only 8 multiplications, that any $a$-digit number times a $b$-digit number will result in an $(a+b)$-digit number. Now lets say out of the given numbers, we have three number such that $a<b<c$, where each variable represents the number of digits in some number. Now we can either multiply $a$ and $b$ together getting an $(a+b)$-digit number which we then multiply by $c$, which if the number of seconds this takes is $S_1$ we would have
$$S_1=ab+(a+b)c=ab+ac+bc$$
Next we could multiply $a$ by $c$ and then multiply that product by $c$; the number of seconds $S_2$ this would take is given by
$$S_2=ac+(a+c)b=ac+ab+bc$$
Lastly we could multiply $b$ and $c$ together, and then multiply that product by $a$; we'll call the time this takes $S_3$ given by
$$S_3=bc+(b+c)a=bc+ab+ac$$
However what we find is that
$$S_1=S_2=S_3$$
And this can be expanded to even more variables and you'll find the same result. So now it's just a matter of finding how long it takes.
Well we have $1$ through $9$ digit numbers, and since if we multiply an $a$-digit number by a $b$-digit number we get an $(a+b)$-digit number, we can expresses the total number of seconds it takes as
$$S=1\cdot 2+(1+2)\cdot 3+(1+2+3)\cdot 4+...+(1+...+8)\cdot 9$$
$$=\sum_{i=1}^{8}\left[\left(\sum_{j=1}^{i}j\right)\cdot (i+1)\right]=\sum_{i=1}^{8}\left[\frac{i(i+1)^2}{2}\right]=870$$
So the minimum number of seconds it will take him is $870$ seconds.
