Question 1182131: I saw this solution online from a tutor (ikelyn). I think there's a mistake in it and i need to make sure I'm right. So in case b in this solution. Shouldn't the equation be 8*8*7*4 = 1792? (4 because there are 4 different possible digits?) In the solution it says 8*8*7 = 448.
The solution:
How many 4-digit even numbers can be formed from the digits 0 to 9
if each digit is to be used only once in each number ?
Solution
The fact that the number is an EVEN number means that the last digit is one of 5 even digits 0, 2, 4, 6, or 8.
In my solution, I will consider two cases separately:
case (a): the last digit is 0 (zero),
and
case (b): the last digit is any of the remaining 4 even digits 2, 4, 6 or 8.
Case (a): the last digit is 0 (zero)
Then the first (most-left) digit is any of 9 remaining digits;
the second digit is any of remaining 8 digits;
the third digit is any of remaining 7 digits.
So, the total number of possible options is 9*8*7 = 504 in this case.
Case (b): the last digit is any of remaining 4 digits 2, 4, 6 or 8.
Then the first (most-left) digit is any of 8 remaining digits (keep in mind that the leading digit CAN NOT be 0 (!));
the second digit is any of 8 remaining digits (zero is ALLOWED in this position);
the third digit is any of 7 remaining digits (zero is ALLOWED in this position).
So, the total number of possible options is 8*8*7 = 448 in this case.
Thus the total number of possibilities is 504 + 448 = 952.
ANSWER. 952 four-digit even numbers can be formed from the digits 0 to 9 if each digit is to be used only once in each number.
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Yes you are correct.
There are 8 choices for slot one, 8 for slot two, 7 for slot three, and 4 choices for the last slot.
So the final answer would be 504+1792 = 2296
Answer by ikleyn(52879)  (Show Source):
You can put this solution on YOUR website! .
Thank you for noticing my error in one of my 1049 lessons in this site.
I just found and fixed it . . .
I am really happy to have so attentive readers.
Thank you again.
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