Question 229152: how many 3 digit number can we do out of these numbers {0,1,2,3,4,5,6}
that are greater than 330.(repetition is allowed)
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
331 to 336 : 6 numbers
340 to 366 : 1 way to choose high digit, 3 ways to choose middle digit, 7 ways to choose low digit : 1 X 3 X 7 = 21
400 to 666 : 3 ways to choose high digit, 7 ways to choose middle digit, 7 ways to choose low digit: 3 X 7 X 7 = 147
6 + 21 + 147 = 174
John
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website! how many 3 digit number can we do out of these numbers {0,1,2,3,4,5,6}
that are greater than 330.(repetition is allowed)
The smallest one allowable is 331.
All must begin with 3,4,5, or 6. However, finding how many begin
with 3 is different from finding how many are langer. Also, of those
beginning with 3, finding how many begin with 33 is different from
finding how many begin with 34, 35, or 36.
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First we count all the three digit numbers beginning with 33:
We can choose only 1, 2, 3, 4, 5, or 6 for the third digit. That's 6
choices for the third digit of those beginning with 33.
So there are 6 allowable 3-digit numbers beginning with 33. These are
331, 332, 333, 334, 335, or 336.
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Next we must find how many begin with 34, 35, or 36.
We can choose the second digit any of 3 ways (either 4, 5 or 6)
We can then choose the third digit any of the total 7 ways.
That's 3x7 or 21 allowable 3-digit numbers beginning with 34, 35, or 36
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Finally we must find how many 3-digit numbers beginning with 4, 5, or 6
We can choose the first digit any of 3 ways (4, 5, or 6)
We can choose the second digit any of the 7 ways.
We can choose the third digit and of the 7 ways.
That's 3x7x7 or 147 allowable 3-digit numbers beginning with 4,5, or 6
Answer: 6 + 21 + 147 = 174
Edwin
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