Question 746487
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<b>Case 1: Repetition allowed</b>


There are ten possible digits, 0 through 9, so there are ten ways to choose the first digit.  Then for each of those ways, since repetition is allowed, there are 10 ways to choose the second digit.  Hence there are 10 times 10 or 100 ways to choose the first two digits.  For each of those 100 ways there are 10 ways to choose the 3rd digit...and so on, so there are *[tex \LARGE 10^5] ways to choose the 5 digits.  Then, for each of those ways there are 26 ways to choose the first letter...and so on.  Altogether: *[tex \LARGE 10^5\,\cdot\,26^2] different combinations.  You can do your own arithmetic.


<b>Case 2: Repetition prohibited</b>


There are ten possible digits, 0 through 9, so there are ten ways to choose the first digit. Then for each of those ways, since repetition is not allowed, there are only 9 ways to choose the second digit since one of the 10 has already been used picking the first digit.  Then there are 8 ways to pick the third digit...and so on.  So there are *[tex \LARGE 10\,\times\,9\,\times\,8\,\times\,7\,\times\,6] ways to pick the five digits, then similarly there are *[tex \LARGE 26\,\times\,25] ways to pick the two letters, so altogether:  *[tex \LARGE 10\,\times\,9\,\times\,8\,\times\,7\,\times\,6\,\times\,26\,\times\,25] different combinations.  Again, you can do the arithmetic yourself.


<b>Super Double Plus Extra Credit -- Case 3: Repetition <i>demanded</i></b>


How many combinations are possible if at least one repetition must occur somewhere in the combination?


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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