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- ☂雨後晴空☀Lv 79 years agoFavorite Answer
2 個不同的非零數字及佔位個數選擇 : 9P2
6個數��列 : 6! / (2! * 4!)
9P2 * 6! / (2! * 4!)= 1080 個可能。
- 地獄之王者-冥王哈帝斯Lv 49 years ago
計算方法: 9P2( 2 個不同的非零數字的可能)x6!/(2!x4!)
(6個數排列)
6!=6x5x4x3x2x1=720
2!x4!=2x1x4x3x2x1=48
9P2 x 6! / (2! x 4!)=1080個不同的可能性
希望我可以幫到你.
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