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
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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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