SOLUTION: How many three-digit numbers can be formed using elements from the set if no element may be used more than once in a number and the number must be odd?

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Question 410233:
How many three-digit numbers can be formed using elements from the set if no element may be used more than once in a number and the number must be odd?
Found 2 solutions by stanbon, sudhanshu_kmr:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
How many three-digit numbers can be formed using elements from the set if no element may be used more than once in a number and the number must be odd?
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Hundreds digit: 9 ways
Tens digit: 9 ways
Ones digit: 5 ways
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Total number: 9*9*5 = 405
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Cheers,
Stan H.
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Answer by sudhanshu_kmr(1152) (Show Source):
You can put this solution on YOUR website!

no. of ways to put an odd digit at unit place = 5 [i.e 1,3,5,7 or 9]

no. of ways to put any other digit at 100's place = 8
( except '0' and that digit which at unit place)

similarly, no. of ways to put any digit at 10's place = 8
( any digit except at unit place and 100's place)

total no. of ways = 5*8*8 = 320