Question 459560
How many 4-digit whole numbers using the digits in the set {1,2,3,4,5,6,7,8,9,0}can be formed using the restrictions below.
For purposes of this problem, the whole number can be "padded" with zeros on the left. So a 0014 is a good number,(IE numbers can start with zeros).
Each part below is is independant of the previous part.
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a. Numbers do not contain the same digit twice, (without replacement)
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Examples would be 0692, 8591

Choose digit #1 10 ways, and then choose digit #2 9 ways, 
and then choose digit #3 8 ways, and then choose digit #4
7 ways.There are 10×9×8×7 = 5040 ways. 
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b. Numbers end with an even digit and can repeat digits, (with replacement)
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Examples would be 8736, 4994, 7770, 0000, 2010

We do not have to choose the digits in any certain order. 
So we will choose digit #4, then digit #1, then digit #2,
and finally digit #3.

Choose digit #4 5 ways (2,4,6,8, or 0), and then choose 
digit #1 10 ways, and then choose digit #2 10 ways, and 
then choose digit #3 10 ways. There are 
5×10×10×10 = 5000 ways.
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c. Numbers begin with an odd digit and can repeat digits, 
(with replacement)
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Examples:  1887, 9742, 5555

Choose digit #1 5 ways (1,3,5,7,9), and then choose 
digit #2 10 ways, and then choose digit #3 10 ways, and 
then choose digit #4 10 ways.  There are 
5×10×10×10 = 5000 ways.
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d. Numbers begin and end with a 1, (with replacement)
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Examples:  1871, 1661, 1111

Choose digit #1 1 way, and then choose digit #2 10 ways, 
and then choose digit #3 10 ways, and then choose digit 
#4 1 way. There are 1×10×10×1 = 100 ways.
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e. Numbers have exactly three digits which are 7s (with replacement)
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Examples: 7771, 7707, 7477, 0777

There are three 7's and one non-7.  We choose the position 
for the non-7 any of 4 ways, and then the digit for the non-7
9 ways, then 1 way each for the 7's.
There are 4×9×1×1 = 36 ways. 

Edwin</pre>