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Homework Statement
"A 3-digit number is made up using the digits 0, 1, 2, 3, 4, 5, 6 and 7 at most once each. The number cannot start with 0. How many such numbers can be formed if:
a. there are no other restrictions
b. the number ends in a 5
c. the number ends in a 0
d. the number is divisible by 5 ?"
2. The attempt at a solution
a:
Three digits: _ _ _
First number can't be 0, 7 other possibilities: 7 _ _
Second number can be 0, 1 number has been used (of the 8), 7 more possibilities: 7 x 7 _
Third number can be any remaining 6: 7 x 7 x 6 = 294 Correct Answer is 180
b:
Three digits: _ _ _
First number can't be 0, 5 is used, 6 other possibilities: 6 _ _
Second number can be 0, 2 numbers used, 6 other possibilities: 6 x 6 _
Third Number is 5: 6 x 6 x 1 = 36 Correct Answer is 25
c:
Three digits: _ _ _
First number can't be 0, 0 is used, 7 other possibilities: 7 _ _
Second number can't be 0, 2 numbers used, 6 other possibilities: 7 x 6 _
Third Number is 0: 7 x 6 x 1 = 42 Correct Answer is 30
d:
Three digits: _ _ _
First number can't be 0, 7 other possibilities: 7 _ _
Second number can be 0, 7 remaining possibilities: 7 x 7 _
Third number is either 0 or 5 (divisible by 5): 7 x 7 x 2 = 98 Correct Answer is 55
I am sorry for this post being so large, but that is all my work. I am probably missing some simple step that is messing up all my answers, however, that is how my book has taught me to look at the problems. Any help with what I'm doing incorrectly would be greatly appreciated!
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