SOLUTION: digits:1,2,3,4,5,6,7,8,9 No repetition is allowed a)how many 5 digit numbers are there that exclude 0 and 7 b) How many of them start with an even number c) how many of them st

Question 843940: digits:1,2,3,4,5,6,7,8,9 No repetition is allowed
a)how many 5 digit numbers are there that exclude 0 and 7
b) How many of them start with an even number
c) how many of them start and end with and even number
Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!
digits:1,2,3,4,5,6,7,8,9 No repetition is allowed
a)how many 5 digit numbers are there that exclude 0 and 7
To exclude 0 and 7 we can only choose from the 8 {1,2,3,4,5,6,8,9} We can choose the 1st digit 8 ways. We can choose the 2nd digit 7 ways. We can choose the 3rd digit 6 ways. We can choose the 4th digit 5 ways. We can choose the 5th digit 4 ways. 8*7*6*5*4 = 6720
b) How many of them start with an even number
We can choose the 1st digit 4 ways. We can choose the 2nd digit 7 ways. We can choose the 3rd digit 6 ways. We can choose the 4th digit 5 ways. We can choose the 5th digit 4 ways. 4*7*6*5*4 = 3360
c) how many of them start and end with and even number.
We can choose the 1st digit 4 ways. We can choose the 5th digit 3 ways. We can choose the 2nd digit 6 ways. We can choose the 3rd digit 5 ways. We can choose the 4th digit 4 ways. 4*3*6*5*4 = 1440 Edwin

Answer by AnlytcPhil(1806) (Show Source): You can put this solution on YOUR website!
digits:1,2,3,4,5,6,7,8,9 No repetition is allowed
a)how many 5 digit numbers are there that exclude 0 and 7
To exclude 0 and 7 we can only choose from the 8 {1,2,3,4,5,6,8,9} We can choose the 1st digit 8 ways. We can choose the 2nd digit 7 ways. We can choose the 3rd digit 6 ways. We can choose the 4th digit 5 ways. We can choose the 5th digit 4 ways. 8*7*6*5*4 = 6720
b) How many of them start with an even number
We can choose the 1st digit 4 ways. We can choose the 2nd digit 7 ways. We can choose the 3rd digit 6 ways. We can choose the 4th digit 5 ways. We can choose the 5th digit 4 ways. 4*7*6*5*4 = 3360
c) how many of them start and end with and even number.
We can choose the 1st digit 4 ways. We can choose the 5th digit 3 ways. We can choose the 2nd digit 6 ways. We can choose the 3rd digit 5 ways. We can choose the 4th digit 4 ways. 4*3*6*5*4 = 1440 Edwin

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