Levi civita tensor contraction help

Problem reads:
Evaluate:
a) dij*djk*dkm*dim
c) Ejk2*Ek2j

The answers in the back of the book for those two are a) 3 , b) 2

My question is, since the kronecker delta is defined for 0 and 1 and the levi-civita symbol is defined for 1, 0, -1, how does one obtain answers that are greater than 1?

You can get a greater then 1 answer because, presumably, the indices are summed over. Looks like they run from 1 to 3.
In a) first sum over m;
dijdjkdk1di1 + dijdjkdk2di2 + dijdjkdk3di3

Next sum each term over i
d1jdjkdk1 + d2jdjkdk2 + d3jdjkdk3

Next sum each term over k
d1jdj1 + d2jdj2 + d3jdj3

Next sum each term over j
1 + 1 + 1 = 3

A little more abstract approach to (a);
I assume all indices i,j,k,m are summed over 1 to 3.
First sum over i;
dmjdjkdkmdmm

Next sum over k;
dmjdjmdmmdmm

Next sum over j
dmmdmmdmmdmm

Next sum over m;
1 + 1 + 1 = 3


I'll let you play with the Levi-civita terms.