Three, (25 points) During the New Year, the grandmother gave three grandsons a total of 400 yuan as New Year's money. There are 50 yuan, 20 yuan, and 10 yuan banknotes available in various quantities for the three grandsons to choose from, but each can only take banknotes of the same denomination. One of them takes a number of banknotes that is exactly equal to the product of the number of banknotes taken by the other two. How many ways are there to choose the denominations and the number of banknotes?
Three people take money in the amounts of $x y$, $x$, and $y$, with denominations of $a$ yuan, $b$ yuan, and $c$ yuan respectively, where $x \geqslant 1, y \geqslant 1$. Then $a x y + b x + c y = 400$.
(1) When $a = b = c$, i.e., the denominations chosen by the three people are the same, then
$$
x y + x + y = \frac{400}{a},
$$

which means $(x+1)(y+1) = \frac{400}{a} + 1$.
(i) If $a = 50$, then
$$
(x+1)(y+1) = 9 = 3 \times 3.
$$

Solving this, we get $x = 2, y = 2$.
Therefore, each of the three people selects 50 yuan in the amounts of 2, 2, and 4 respectively. In this case, there is one way to choose.
(ii) If $a = 20$, then
$$
(x+1)(y+1) = 21 = 3 \times 7.
$$

Solving this, we get $x = 2, y = 6$.
Therefore, there is one way to choose 20 yuan.
(iii) If $a = 10$, then $(x+1)(y+1) = 41$. In this case, $x$ and $y$ have no positive integer solutions.
(2) When two of $a$, $b$, and $c$ are equal, based on the lack of order in selection, it can be divided into two cases: $a = b \neq c$ and $b = c \neq a$.
$$
\begin{array}{l}
\text{When } a = b \neq c, \\
a x y + a x + c y = 400,
\end{array}
$$

which means $(a x + c)(y + 1) = 400 + c(y + 1 \geqslant 2)$.
(i) If $a = 50, c = 20$, then
$$
(50 x + 20)(y + 1) = 420,
$$

which means $(5 x + 2)(y + 1) = 42 = 7 \times 6$.
Solving this, we get $x = 1, y = 5$.
In this case, there is one way to choose.
(ii) If $a = 50, c = 10$, then
$(50 x + 10)(y + 1) = 410$,

which means $(5 x + 1)(y + 1) = 41$ (a prime number).
The above equation has no positive integer solutions.
(iii) If $a = 20, c = 50$, then
$(20 x + 50)(y + 1) = 450$,

which means $(2 x + 5)(y + 1) = 45 = 9 \times 5 = 15 \times 3$.
Solving this, we get $x = 2, y = 4$ or $x = 5, y = 2$.
Therefore, there are two ways to choose the denominations and amounts.
Similarly, we get:
(iv) If $a = 20, c = 10$, there are no positive integer solutions.
(v) If $a = 10, c = 50$, there are two ways to choose the denominations and amounts.
(vi) If $a = 10, c = 20$, there are five ways to choose the denominations and amounts.
(3) When $a$, $b$, and $c$ are all different, similarly, we get:
(i) If $a = 10, b = 20, c = 50$, there is one way to choose the denominations and amounts.
(ii) If $a = 20, b = 10, c = 50$, there is one way to choose the denominations and amounts.
(iii) If $a = 50, b = 10, c = 20$, there are no positive integer solutions.
In summary, there are 14 ways to choose the denominations and amounts.
(Xie Wenxiao, Huanggang High School, Hubei, 438000)