Question

Given the digits 5, 4, 2, 0, 3, 7, 9, 6, 8. How many three-digit number codes can be formed if the hundreds
place must be even and the ones place is at least 5? No repition

Answer 1

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ones place must have at-least 5: it can have 5,6,7,8,9 = 5 ways

2 cases arise:

Case1:

If it is an even number (6,8 - 2 ways) - then the 100s digit can be filled in 3 ways ( 0 can't be put)and then the 2nd blank in 10s can have 8 unique numbers.

So, 3*8*2 = 48 ways

Case2:

If it is an odd number (5,7,9 - 3 ways) - then the 100s digit can be filled in 4 ways ( 0 can't be put), then 2nd blank in 10s can have 8 unique numbers.

So, 4*8*3 = 96 ways

So, total ways in which three-digit number codes can be formed if the hundreds place must be even and the ones place is at least 5 with no repetation is 144 ways