Solution:
Let the number of pairs of balls, clubs, and hats purchased be $b$, $c$, $h$ respectively. Then, we must have

$$\text{cost:}\qquad \frac{1}{4} b + h + 10c = 100$$
$$\text{quantity:}\qquad 2b + h + c = 100$$

Subtracting these yields $\frac{7}{4} b - 9c = 0$ and thus
$$7b = 36c.$$
Since $\gcd(7,36) = 1$ this implies $36 \mid b$. Since $0 < b < 51$, we must have $b = 36$. Therefore $c = 7$. Thus $h = 100 - 2 \times 36 - 7 = 21$. Hence the final answer is $(b, h, c) = (36, 21, 7)$. Checking we get:

$$\text{cost:}\qquad \frac{1}{4} b + h + 10c = 9 + 21 + 70 = 100$$
$$\text{quantity:}\qquad 2b + h + c = 72 + 21 + 7 = 100.$$