Problem 1627

To solve the problem, we need to count the number of ordered quadruples $(a, b, c, d)$ such that $1 \le a < b < c < d \le 10$ and $a + d > b + c$. We can reframe the condition $a + d > b + c$ as $d - c > b - a$. We will break this into cases based on the value of $d - c$.

### Case 1: $d - c = 2$
For $d - c = 2$, we need $b - a < 2$. This means $b - a = 1$.

- If $(a, b) = (1, 2)$, then $c$ and $d$ can be $(3, 5), (4, 6), (5, 7), (6, 8), (7, 9), (8, 10)$. This gives 6 pairs.
- If $(a, b) = (2, 3)$, then $c$ and $d$ can be $(4, 6), (5, 7), (6, 8), (7, 9), (8, 10)$. This gives 5 pairs.
- If $(a, b) = (3, 4)$, then $c$ and $d$ can be $(5, 7), (6, 8), (7, 9), (8, 10)$. This gives 4 pairs.
- If $(a, b) = (4, 5)$, then $c$ and $d$ can be $(6, 8), (7, 9), (8, 10)$. This gives 3 pairs.
- If $(a, b) = (5, 6)$, then $c$ and $d$ can be $(7, 9), (8, 10)$. This gives 2 pairs.
- If $(a, b) = (6, 7)$, then $c$ and $d$ can be $(8, 10)$. This gives 1 pair.

Total for $d - c = 2$ is $6 + 5 + 4 + 3 + 2 + 1 = 21$.

### Case 2: $d - c = 3$
For $d - c = 3$, we need $b - a < 3$. This means $b - a = 1$ or $b - a = 2$.

- For $b - a = 1$:
- If $(a, b) = (1, 2)$, then $c$ and $d$ can be $(3, 6), (4, 7), (5, 8), (6, 9), (7, 10)$. This gives 5 pairs.
- If $(a, b) = (2, 3)$, then $c$ and $d$ can be $(4, 7), (5, 8), (6, 9), (7, 10)$. This gives 4 pairs.
- If $(a, b) = (3, 4)$, then $c$ and $d$ can be $(5, 8), (6, 9), (7, 10)$. This gives 3 pairs.
- If $(a, b) = (4, 5)$, then $c$ and $d$ can be $(6, 9), (7, 10)$. This gives 2 pairs.
- If $(a, b) = (5, 6)$, then $c$ and $d$ can be $(7, 10)$. This gives 1 pair.

Total for $b - a = 1$ is $5 + 4 + 3 + 2 + 1 = 15$.

- For $b - a = 2$:
- If $(a, b) = (1, 3)$, then $c$ and $d$ can be $(4, 7), (5, 8), (6, 9), (7, 10)$. This gives 4 pairs.
- If $(a, b) = (2, 4)$, then $c$ and $d$ can be $(5, 8), (6, 9), (7, 10)$. This gives 3 pairs.
- If $(a, b) = (3, 5)$, then $c$ and $d$ can be $(6, 9), (7, 10)$. This gives 2 pairs.
- If $(a, b) = (4, 6)$, then $c$ and $d$ can be $(7, 10)$. This gives 1 pair.

Total for $b - a = 2$ is $4 + 3 + 2 + 1 = 10$.

Total for $d - c = 3$ is $15 + 10 = 25$.

### Case 3: $d - c = 4$
For $d - c = 4$, we need $b - a < 4$. This means $b - a = 1, 2, 3$.

- For $b - a = 1$:
- If $(a, b) = (1, 2)$, then $c$ and $d$ can be $(3, 7), (4, 8), (5, 9), (6, 10)$. This gives 4 pairs.
- If $(a, b) = (2, 3)$, then $c$ and $d$ can be $(4, 8), (5, 9), (6, 10)$. This gives 3 pairs.
- If $(a, b) = (3, 4)$, then $c$ and $d$ can be $(5, 9), (6, 10)$. This gives 2 pairs.
- If $(a, b) = (4, 5)$, then $c$ and $d$ can be $(6, 10)$. This gives 1 pair.

Total for $b - a = 1$ is $4 + 3 + 2 + 1 = 10$.

- For $b - a = 2$:
- If $(a, b) = (1, 3)$, then $c$ and $d$ can be $(4, 8), (5, 9), (6, 10)$. This gives 3 pairs.
- If $(a, b) = (2, 4)$, then $c$ and $d$ can be $(5, 9), (6, 10)$. This gives 2 pairs.
- If $(a, b) = (3, 5)$, then $c$ and $d$ can be $(6, 10)$. This gives 1 pair.

Total for $b - a = 2$ is $3 + 2 + 1 = 6$.

- For $b - a = 3$:
- If $(a, b) = (1, 4)$, then $c$ and $d$ can be $(5, 9), (6, 10)$. This gives 2 pairs.
- If $(a, b) = (2, 5)$, then $c$ and $d$ can be $(6, 10)$. This gives 1 pair.

Total for $b - a = 3$ is $2 + 1 = 3$.

Total for $d - c = 4$ is $10 + 6 + 3 = 19$.

### Case 4: $d - c = 5$
For $d - c = 5$, we need $b - a < 5$. This means $b - a = 1, 2, 3, 4$.

- For $b - a = 1$:
- If $(a, b) = (1, 2)$, then $c$ and $d$ can be $(3, 8), (4, 9), (5, 10)$. This gives 3 pairs.
- If $(a, b) = (2, 3)$, then $c$ and $d$ can be $(4, 9), (5, 10)$. This gives 2 pairs.
- If $(a, b) = (3, 4)$, then $c$ and $d$ can be $(5, 10)$. This gives 1 pair.

Total for $b - a = 1$ is $3 + 2 + 1 = 6$.

- For $b - a = 2$:
- If $(a, b) = (1, 3)$, then $c$ and $d$ can be $(4, 9), (5, 10)$. This gives 2 pairs.
- If $(a, b) = (2, 4)$, then $c$ and $d$ can be $(5, 10)$. This gives 1 pair.

Total for $b - a = 2$ is $2 + 1 = 3$.

- For $b - a = 3$:
- If $(a, b) = (1, 4)$, then $c$ and $d$ can be $(5, 10)$. This gives 1 pair.

Total for $b - a = 3$ is $1$.

Total for $d - c = 5$ is $6 + 3 + 1 = 10$.

### Case 5: $d - c = 6$
For $d - c = 6$, we need $b - a < 6$. This means $b - a = 1, 2, 3, 4, 5$.

- For $b - a = 1$:
- If $(a, b) = (1, 2)$, then $c$ and $d$ can be $(3, 9), (4, 10)$. This gives 2 pairs.
- If $(a, b) = (2, 3)$, then $c$ and $d$ can be $(4, 10)$. This gives 1 pair.

Total for $b - a = 1$ is $2 + 1 = 3$.

- For $b - a = 2$:
- If $(a, b) = (1, 3)$, then $c$ and $d$ can be $(4, 10)$. This gives 1 pair.

Total for $b - a = 2$ is $1$.

Total for $d - c = 6$ is $3 + 1 = 4$.

### Case 6: $d - c = 7$
For $d - c = 7$, we need $b - a < 7$. This means $b - a = 1, 2, 3, 4, 5, 6$.

- For $b - a = 1$:
- If $(a, b) = (1, 2)$, then $c$ and $d$ can be $(3, 10)$. This gives 1 pair.

Total for $d - c = 7$ is $1$.

### Summing Up All Cases
Adding up all the cases, we get:
$$21 + 25 + 19 + 10 + 4 + 1 = 80$$

The final answer is $\boxed{80}$