Question 891210: How many four-digit even numbers can be formed from the digits 0,2,3,5,6, and 9 if all for digits are different? The given answer is 108, but when i tried it out i got stuck because of the extra number and got 120 .....
Found 2 solutions by AnlytcPhil, richwmiller:
Answer by AnlytcPhil(1806)   (Show Source):
You can put this solution on YOUR website!
There are two cases since 0 is an even digit and it can go last, but it
can't go first. So the number of choices for the first digit is different
when we choose 0 for the last digit than if we choose 2 or 6 for the last
digit. So we must break this problem into two cases:
Case 1: 0 goes last
Choose the last (5th) digit as 0 in 1 way.
Choose the 1st digit any of 5 ways. {2,3,5,6,9}
Choose the 2nd digit any of the remaining 4 ways.
Choose the 3rd digit any of the remaining 3 ways.
Choose the 4th digit either of the remaining 2 ways.
That's 1×5×4×3×2 = 120 ways ending in 0
Case 2: 2 or 6 goes last:
Choose the last (5th) digit as in either of 2 ways.
Choose the 1st digit any of 4 ways. (CANNOT choose 0)
Choose the 2nd digit any of the remaining 4 ways. (CAN choose 0)
Choose the 3rd digit any of the remaining 3 ways.
Choose the 4th digit either of the remaining 2 ways.
That's 2×4×4×3×2 = 192 ways ending in 2 or 6
Total for the 2 cases = 120+192 = 312
Edwin
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! Edwin gave you the count for five digit even numbers. We want four digit even numbers
120/2=60
192/4=48
1×5×4×3=60
2×4×3×2=48
60+48=108
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