\(2700 = 2^2 \times 3^3 \times 5^2\)
因為 \(a,b,c\) 都是 \(2700\) 的正因數,
所以,可令 \(a=2^{x_1}3^{y_1}5^{z_1},b=2^{x_2}3^{y_2}5^{z_2},c=2^{x_3}3^{y_3}5^{z_3}\),
其中 \(x_1,x_2,x_3\in\left\{0,1,2\right\}\),\(y_1,y_2,y_3\in\left\{0,1,2,3\right\}\),\(z_1,z_2,z_3\in\left\{0,1,2\right\}\)
先來算題目要求之機率的分母=(給你算,我不要算
)
再來算分子:
因為題目說 \(a\Big|b\)、\(b\Big|c\),
所以 \(0\leq x_1\leq x_2\leq x_3\leq 2\),\(0\leq y_1\leq y_2\leq y_3\leq 3\),\(0\leq z_1\leq z_2\leq z_3\leq 2\),
故題目要求之機率的分子為 \(H^4_2 \times H^4_3 \times H^4_2 = C^5_2 C^6_3 C^4_2 = 2000.\)
或是 \(H^3_3\times H^4_3 \times H^3_3=C^5_3 C^6_3 C^5_3 = 2000.\)
(最後那兩個式子,分別對應於下面這篇文章的兩個方法:
https://math.pro/db/thread-63-1-1.html)
所以題目所求之機率 \(\displaystyle =\frac{2000}{3^3 4^3 3^3} = \frac{125}{2916}.\)
出處:99台中一中教甄