Problem 1240

1. We need to find the average value of the fourth element $d$ in the set $\mathcal{S} = \{(a, b, c, d, e) : 0 < a < b < c < d < e < 100\}$, where $a, b, c, d, e$ are integers.
2. First, we determine the total number of such tuples. This is given by the binomial coefficient $\binom{99}{5}$, since we are choosing 5 distinct integers from the set $\{1, 2, \ldots, 99\}$.
3. Next, we need to calculate the sum of all possible values of $d$ over all tuples in $\mathcal{S}$. We will use the fact that $d$ can take any value from 4 to 98, and for each value of $d$, we need to count how many tuples have $d$ as the fourth element.
4. For a fixed $d$, the number of ways to choose $a, b, c$ from $\{1, 2, \ldots, d-1\}$ is $\binom{d-1}{3}$, and the number of ways to choose $e$ from $\{d+1, d+2, \ldots, 99\}$ is $99 - d$.
5. Therefore, the total number of tuples with $d$ as the fourth element is $\binom{d-1}{3} \cdot (99 - d)$.
6. The sum of all values of $d$ weighted by the number of tuples is:
$$\sum_{d=4}^{98} d \cdot \binom{d-1}{3} \cdot (99 - d)$$
7. To find the average value $D$, we divide this sum by the total number of tuples $\binom{99}{5}$:
$$D = \frac{\sum_{d=4}^{98} d \cdot \binom{d-1}{3} \cdot (99 - d)}{\binom{99}{5}}$$
8. We simplify the numerator using the identity for binomial coefficients:
$$\sum_{d=4}^{98} d \cdot \binom{d-1}{3} \cdot (99 - d) = 4 \sum_{i=1}^{95} i \cdot \binom{99-i}{4}$$
9. Using the hockey-stick identity for binomial coefficients, we have:
$$\sum_{i=1}^{95} i \cdot \binom{99-i}{4} = \binom{100}{6}$$
10. Therefore, the numerator becomes:
$$4 \cdot \binom{100}{6}$$
11. The denominator is $\binom{99}{5}$, so:
$$D = \frac{4 \cdot \binom{100}{6}}{\binom{99}{5}}$$
12. Using the property of binomial coefficients:
$$\binom{100}{6} = \frac{100!}{6! \cdot 94!} \quad \text{and} \quad \binom{99}{5} = \frac{99!}{5! \cdot 94!}$$
13. Simplifying the ratio:
$$\frac{\binom{100}{6}}{\binom{99}{5}} = \frac{100 \cdot 99}{6 \cdot 5} = \frac{100 \cdot 99}{30} = \frac{9900}{30} = 330$$
14. Thus:
$$D = \frac{4 \cdot 330}{1} = 1320$$
15. Finally, the largest integer less than or equal to $D$ is:
$$\boxed{66}$$