Thanks for pointing out my ignorance of the associated probabilities,
The solution I get now appears complex to me...and I would like to take your views on simplifying it in case it is correct,
Probability of selecting permutation with sum 1:
$P_{1} = n! * \sum_{i = 1}^{n} p_{i} \prod_{j = 1, j \ne i}^{n} (1-p_{j})$
Here we have $n!$ because we choose random permutation of elements in A.
Similarly for sum 2:
$P_{2} = n! * \sum_{i_{1} = 1}^{n}\sum_{i_{2} = 1, i_{1} \ne i_{2}}^{n} p_{i_{1}} p_{i_{2}}\prod_{j = 1, j \ne i_{1}, j \ne i_{2}}^{n} (1-p_{j})$
Generalizing it for sum n we get:
$P_{n} = n! * p_{i_{1}}* p_{i_{2}}...*p_{i_{n}}$, since there is only one string on n 1s, although it still can be arranged in $n!$ ways because all elements are distinct.
So the expected sum = $\sum_{i = 1}^{n} i * P_{i}$
