To calculate the theoretical probability that applesauce and broccoli will both be offered among the three chosen side dishes from the six available options, we need to first determine the total number of ways to choose three dishes from the six available dishes.
The total number of combinations of choosing 3 from 6 dishes is given by the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n - r)!} \]
where \( n \) is the total number of items (dishes), and \( r \) is the number of items to choose.
In this case, \( n = 6 \) and \( r = 3 \):
\[ \binom{6}{3} = \frac{6!}{3!(6 - 3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
Next, we need to find the number of successful outcomes where both applesauce and broccoli are included in the selection of three dishes. If applesauce and broccoli are included, then we need to choose 1 more dish from the remaining 4 dishes (corn, dumplings, egg rolls, French fries).
The number of ways to choose 1 dish from these 4 remaining dishes is:
\[ \binom{4}{1} = 4 \]
Now, we calculate the probability. The probability \( P \) of both applesauce and broccoli being present in the selection is given by the number of successful outcomes divided by the total number of outcomes:
\[ P(\text{applesauce and broccoli}) = \frac{\text{Number of successful outcomes}}{\text{Total outcomes}} = \frac{4}{20} = \frac{1}{5} = 0.20 \]
Thus, the theoretical probability that applesauce and broccoli will both be offered on Monday is:
\[ \boxed{20%} \]