SOLUTION: 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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Question 1106687: 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
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


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, there are 10 choices for the tens digit after the units digit is chosen. The total number of possible 2-digit numbers is 10*5 = 50.

(b) If repetition is not allowed, there are only 9 choices left for the tens digit after the units digit is chosen. The total number of possible 2-digit numbers is 9*5 = 45.

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Oops! I didn't spend enough time thinking about this one....

The tens digit can't be 0....

(a) If repetition is allowed, we have 5 choices for the units digit (any of the 5 even digits); then we have 9 choices for the tens digit (any digit except 0). That makes 5*9 = 45 2-digit numbers.

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

If the units digit is 0, then we still have 9 digits to choose from for the tens digit; that makes 9 possible 2-digit numbers with units digit 0.

If the units digit is any of the other 4 even digits, then we have only 8 choices for the tens digit (it again can't be 0; and it also can't be the same as the units digit); that makes 4*8=32 2-digit numbers with units digit 2, 4, 6, or 8.

So with repetition not allowed, the total number of possible 2-digit numbers is 9+32=41.
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 


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,  if repetition is allowed.



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


    Case 1.  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".



    Case 2.  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  case 1 + case 2  is  9 + 32 = 41.


             So, in this problem / (sub-problem) the answer is:  41  2-digit even numbers are possible,  if repetition is not allowed.

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Notice that the answers to  (a)  and  (b)  are  CONSISTENT: 
    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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