Discussion
A box contains 24 identical balls of which one ball is black and the remaining balls are green. Three balls are taken simultaneously and randomly. The number of ways of getting only green balls, is
Answer: D
The number of ways is 1771.
Getting only green balls means choosing all three from the 23 green balls (the balls are treated as distinguishable when counting selections):
\[ \begin{aligned} {}^{23}C_{3} &= \frac{23 \times 22 \times 21}{3 \times 2 \times 1} \\ &= 1771 \end{aligned} \]
As a check, all selections number \({}^{24}C_{3} = 2024\), and those including the black ball number \({}^{23}C_{2} = 253\); \(2024 - 253 = 1771\). The nearby values 1763 to 1765 are arithmetic slips.
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