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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 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 . // 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.
