atulcse
asked
in Combinatory
Jan 12

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## 1 Answer

1 vote

Number of 5-digit even numbers with all distinct digits

Case 1 : Fix 0 at one’s place and most significant digit can’t be 0

So number of possibilities = **9 X 8 X 7 X 6 X1** = **3024**

Case 2 : We can have {2,4,6,8} at one’s place(hence 4 possibility) and most significant digit can’t be 0

So number of possibilities = **8 X 8 X 7 X 6 X4** = **10752**

Total = **13776**