B2. How many quadruples of natural numbers $(a, b, c, d)$ satisfy the inequalities

$$
a>b>c>d \quad \text{and} \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}>2 ?
$$
B2. If $d \geq 2$, then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \leq 4 \cdot \frac{1}{2}=2$, which is a contradiction. Therefore, $d=1$. If $c \geq 3$, then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \leq 3 \cdot \frac{1}{3}+1=2$, which is again a contradiction. Therefore, $c=2$. Similarly, it follows that $b=3$, otherwise we would have $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \leq 2 \cdot \frac{1}{4}+\frac{1}{2}+1=2$. Thus, $\frac{1}{a}+\frac{1}{3}+\frac{1}{2}+\frac{1}{1}>2$, or $\frac{1}{a}>\frac{1}{6}$. Since $a>b=3$, $a$ can only be 4 or 5. The inequality is satisfied by two quadruples, $(4,3,2,1)$ and $(5,3,2,1)$.

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-21.jpg?height=69&width=1582&top_left_y=525&top_left_x=274)

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-21.jpg?height=54&width=1579&top_left_y=584&top_left_x=273)

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-21.jpg?height=51&width=1576&top_left_y=637&top_left_x=274)

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-21.jpg?height=66&width=1576&top_left_y=681&top_left_x=274)

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-21.jpg?height=57&width=1579&top_left_y=734&top_left_x=273)
(If the contestant simply writes down both solutions, only 2 points are awarded.)