Using only 0 , 1 , 2 , 3 ,4 how many three digit
(a) even number can be formed (with no repetition ) ?
Case 1: First number even
1. We choose the first digit even in 2 ways, (as 2 or 4, not 0).
2. We choose the last digit even in 2 ways, (either of the remaining 2 unused
even digits.
3. We choose the middle digit as any of the remaining 3 digits.
That's 2x2x3 = 12 ways for case 1.
Case 2: First digit odd.
1. We choose the first digit odd in 2 ways, (as 1 or 3)
2. We choose the last digit even in 3 ways, (as 0 or 2 of 4)
3. We choose the middle digit as any of the remaining 3 digits.
That's 2x3x3 = 18 ways for case 2.
Total even numbers: 12+18 = 30
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(b) odd number can be formed (with no repetition) ?
Case 1: First number even
1. We choose the first digit even in 2 ways, (as 2 or 4, not 0).
2. We choose the last digit odd in 2 ways, (as 1 or 3)
3. We choose the middle digit as any of the remaining 3 digits.
That's 2x2x3 = 12 ways for case 1.
Case 2: First digit odd.
1. We choose the first digit odd in 2 ways, (as 1 or 3)
2. We choose the last digit odd in the 1 remaining way.
3. We choose the middle digit as any of the remaining 3 digits.
That's 2x1x3 = 6 ways for case 2.
Total odd numbers: 12+6 = 18
Edwin