Summing Conditional Products

since we have to find the limit when n goes to infinity, we need only consider the highest degree term of the numerator.
had the condition a<b<c<d<e not been imposed the sum would have been.
${(1+2+3..n)}^{5}$ . (notice when opened this gets all the five-tuples of integers less than n). = ${(n(n+1))}^{5}$ / $2^{5}$ .
for n goes to infinity when the above expression is expanded most of the tuples will be repeated 5!=120 times. the number of terms not repeated will be of lower order.
thus the $n^{10}$ term of the numerator is $n^{10}$ /(32*120).
thusA/B=1/3840.

This is not an interesting solution but I think it is natural to think about.

Just similar to do multivariable intergral. We sum over the range of 1 variable and keep other variables unchanged. (For multivariable calculus, we care about limit of intergration. Here we care about the summing range of variables ) Do the summation 5 times, we get:

$\frac{n (1 + n) (-240 n + 292 n^2 + 1828 n^3 + 2227 n^4 + 1240 n^5 + 358 n^6 + 52 n^7 + 3 n^8)}{11520 n^{10}}$

Clearly, we get

$A=1$ and $B=3840$ .

Hence,

$A+B=\boxed{3841}$

we can also use newton's sums to get a general expression in n and then apply limits,or we can apply limits when we think we are close to the answer,we actually don't have to get the entire expression