Permutation & combination
Asked by | 29th Nov, 2009, 02:45: PM
We can select only the first and last digit and permute the remaining three numbers,
1/ First digit 1 and the last digit can be 0, 2 or 4. For each last digit we have to permute 3 remaining digits. This gives a total of 3!x3 = 18 numbers. For example, 10324.
2/With first digit as 2, the last digit can be 0 and 4 only. This gives 3!x2 = 12 numbers.
3/With first digit as 3, the last digit can be 0, 2 and 4. This gives 3!x3 = 18 numbers.
4/With first digit as 4, the last digit can be 0 and 2. This gives 3!x2 = 12 numbers.
Hence the total even numbers are 18+12+18+12 = 60
Answered by | 29th Nov, 2009, 05:28: PM
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