Permutation
回答 (2)
2010-11-25 5:17 am
✔ 最佳答案
For no E there are 3!=6 permutations (B,T,L)For 1 E, there are C(3,2)=3 choices for the rest 2 letters.There are 3×3!=18 permutationsFor 2 E’s, there are 3 choices for the rest 1 letterThere are 3×3=9 permutationsFor 3 E' s there is only 1 permutationTotal number of permutations=6+18+9+1=34 
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