Question 1106687
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How many two-digit even numbers can you form using the digits 0, 1 , 2, 3, 4, 5, 6, 7, 8, 9?
a. repetition is allowed
b. no repetition
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<pre>
Of the 10 digits, 5 are even.  So in both cases there are 5 choices for the units digit of the 2-digit number.


(a) If repetition is allowed, then there are 10-1 = 9 choices for the tens digit.  

    (9 because you can not use zero as the "tens" digit).

    The total number of possible 2-digit even numbers is 9*5 = 45,  <U>if repetition is allowed</U>.



(b) If repetition is not allowed, then there are two cases.


    <U>Case 1.</U>  The "units" digit is zero. Then you have 9 choices for the "tens" digits.

             Hence, 9 (nine)  2-digit numbers are possible, ending by "0".



    <U>Case 2.</U>  The "units" digit is any of 4 even digits 2, 4, 6 or 8. 

             Then you have 8 choices for the "tens" digit.

                  (Only 8, because you can not use zero and just used "units" digit).

             Hence, 8*4 = 32  2-digit numbers are possible in this case.

             The total  for  <U>case 1 + case 2</U>  is  9 + 32 = 41.


             So, in this problem / (sub-problem) the answer is:  41  2-digit even numbers are possible,  <U>if repetition is not allowed</U>.
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Notice that the answers to &nbsp;(a) &nbsp;and &nbsp;(b) &nbsp;are &nbsp;<U>CONSISTENT</U>:

<pre>
    From 45 numbers of the set (a) you need exclude four numbers 22, 44, 66  and 88  to get  41 number of the set (b).
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