Problem 1590

To solve the problem, we need to find the number of ordered quadruples $(a, b, c, d)$ of distinct positive integers such that $\binom{\binom{a}{b}}{\binom{c}{d}} = 21$.

First, we note that $21$ can be expressed as a binomial coefficient in the following ways:
$$21 = \binom{7}{2} = \binom{7}{5} = \binom{21}{1} = \binom{21}{20}$$

We will consider each case separately and count the number of valid quadruples $(a, b, c, d)$ for each case.

1. **Case 1: $\binom{\binom{a}{b}}{\binom{c}{d}} = \binom{7}{2}$**
- $\binom{a}{b} = 7$
- $\binom{c}{d} = 2$

To make $\binom{a}{b} = 7$, we have:
$$\binom{7}{1} = 7 \quad \text{and} \quad \binom{7}{6} = 7$$

To make $\binom{c}{d} = 2$, we have:
$$\binom{2}{1} = 2$$

Therefore, the possible quadruples are:
$$(7, 1, 2, 1) \quad \text{and} \quad (7, 6, 2, 1)$$

However, the numbers must be distinct, so $(7, 1, 2, 1)$ is not valid. Thus, there is only $1$ valid solution:
$$(7, 6, 2, 1)$$

2. **Case 2: $\binom{\binom{a}{b}}{\binom{c}{d}} = \binom{7}{5}$**
- $\binom{a}{b} = 7$
- $\binom{c}{d} = 5$

To make $\binom{a}{b} = 7$, we have:
$$\binom{7}{1} = 7 \quad \text{and} \quad \binom{7}{6} = 7$$

To make $\binom{c}{d} = 5$, we have:
$$\binom{5}{1} = 5 \quad \text{and} \quad \binom{5}{4} = 5$$

Therefore, the possible quadruples are:
$$(7, 1, 5, 1), (7, 1, 5, 4), (7, 6, 5, 1), (7, 6, 5, 4)$$

However, the numbers must be distinct, so $(7, 1, 5, 1)$ is not valid. Thus, there are $3$ valid solutions:
$$(7, 1, 5, 4), (7, 6, 5, 1), (7, 6, 5, 4)$$

3. **Case 3: $\binom{\binom{a}{b}}{\binom{c}{d}} = \binom{21}{1}$**
- $\binom{a}{b} = 21$
- $\binom{c}{d} = 1$

There are no ways to make $\binom{c}{d} = 1$ with distinct positive integers, so there are no valid solutions for this case.

4. **Case 4: $\binom{\binom{a}{b}}{\binom{c}{d}} = \binom{21}{20}$**
- $\binom{a}{b} = 21$
- $\binom{c}{d} = 20$

To make $\binom{a}{b} = 21$, we have:
$$\binom{7}{2} = 21 \quad \text{and} \quad \binom{7}{5} = 21 \quad \text{and} \quad \binom{21}{1} = 21 \quad \text{and} \quad \binom{21}{20} = 21$$

To make $\binom{c}{d} = 20$, we have:
$$\binom{20}{1} = 20 \quad \text{and} \quad \binom{20}{19} = 20 \quad \text{and} \quad \binom{6}{3} = 20$$

Therefore, the possible quadruples are:
$$(21, 1, 20, 1), (21, 1, 20, 19), (21, 1, 6, 3), (21, 20, 20, 1), (21, 20, 20, 19), (21, 20, 6, 3), (7, 2, 20, 1), (7, 2, 20, 19), (7, 2, 6, 3), (7, 5, 20, 1), (7, 5, 20, 19), (7, 5, 6, 3)$$

However, the numbers must be distinct, so $(21, 1, 20, 1)$, $(21, 20, 20, 1)$, and $(21, 20, 20, 19)$ are not valid. Thus, there are $9$ valid solutions:
$$(21, 1, 20, 19), (21, 1, 6, 3), (21, 20, 6, 3), (7, 2, 20, 1), (7, 2, 20, 19), (7, 2, 6, 3), (7, 5, 20, 1), (7, 5, 20, 19), (7, 5, 6, 3)$$

Adding up all the valid solutions from each case, we get:
$$1 + 3 + 0 + 9 = 13$$

The final answer is $\boxed{13}$.