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分類討論, 當有兩個2時, 第一二位數為2的組合有2X3=6種 第二三位數為2的組合有2X2=4種 第三四位數為2的組合有2X2=4種 當有三個2時, 有第一二三位數為2或者是第二三四位數為2兩種可能 當有四個2時,有一種可能 所以至少有連續兩位是2的有6+4+4+2+1=17個
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