SOLUTION: how many numbers can you form using the digits 2, 3, 4, 6, 5, 1, 0 if the numbers formed are even numbers with non-repeating digits?

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Question 1118501: how many numbers can you form using the digits 2, 3, 4, 6, 5, 1, 0 if the numbers formed are even numbers with non-repeating digits?
Found 3 solutions by greenestamps, solver91311, ikleyn:
Answer by greenestamps(13203) About Me  (Show Source):
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1-digit numbers: 4 (0, 2, 4, and 6)
2-digit numbers: 4*6 = 24 (choice of 4 digits for the last digit; choice of the 6 others for the first)
3-digit numbers: 4*6*5 = 120
4-digit numbers: 4*6*5*4 = 480
5-digit numbers: 4*6*5*4*3 = 1440
6-digit numbers: 4*6*5*4*3*2 = 2880
7-digit numbers: 4*6*5*4*3*2*1 = 2880

total: 7828

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


There are four possibilities for the units digit. For each of these possibilities, there remain 6 ways to choose the 10s digit, so 24 ways to choose the 10s and units digits. For each of these 24 possibilities, there are 5 ways to choose the 100s digit. And so on.



You can do your own arithmetic


John

My calculator said it, I believe it, that settles it

Answer by ikleyn(52855) About Me  (Show Source):
You can put this solution on YOUR website!
.
how many numbers can you form using the digits 2, 3, 4, 6, 5, 1, 0 if the numbers formed are even numbers with non-repeating digits?
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Let me edit your formulation in this way 

    how many highlight%287-digit%29 numbers can you form using the digits 2, 3, 4, 6, 5, 1, 0 
    if the numbers formed are even numbers with non-repeating digits?

by adding the word/the words highlight%287-digit%29.   //   I think it is exactly what you were going to ask.


Solution


Case 1:  The last digit is 0 (zero) 

Then the number of possibilities (of ways) is  6*5*4*3*2*1 = 720.


Case 2:  the last digit is 2, or 4, or 6. 

Then, since "0" can not be the leading digit, there are only 5 possibilities for the leading digit. 


Moving from the second digit from the left to the sixth from the left, we have consequently 5, 4, 3, 2, 1 possibilities, that give, in all,

    (5*5*4*3*2*1)*3 =  1800 ways.



Finally, adding the numbers 720 and 1800, you get the total number of the possible numbers under the question 2520.

Answer.   2520 numbers.

Any other answer is incorrect,  so ignore it.


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The lesson to learn from my solution: 

    If you are asked to form a number/numbers from given digits, and if there is "0" (zero) among the given digits,

    then be very CAREFUL and REMEMBER that 0 is very special digit: it can not go first and can not be a leading digit.