document.write( "Question 230672: How many 4-digit number are even of the number set ( 0,1,2,3,4,5) if 0151 is not considered a 4-digit \n" ); document.write( "

Algebra.Com's Answer #170849 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
(NOTE: This solution includes a correction to the one provided earlier. My earlier solution mistakenly included 6 among the possible digits. However the new answer still does not match the proposed solution of 456. If the problem is to find the number of all possible even, 4-digit numbers which do not start with zero and which are formed only using 0, 1, 2, 3, 4, 5 as digits, then I believe my solution is correct.
\n" ); document.write( "The only ways 456 can be correct is if I have misunderstood the problem or if there are additional conditions which are not specified.
\n" ); document.write( "In the message I received from the poster, there is a statement: \"It has to be a two case solution.\" This is followed by \"One is to factor the zero in the last position and then case 2 is to take the remaining choices with other 2 even ex zero.\" which makes no sense to me. There is no reason I can see, given the problem as stated above, to consider a zero in the last (ones) position as a separate case.)

\n" ); document.write( "The conditions:

Digits are from the set {0, 1, 2, 3, 4, 5}
4 digits (without a leading 0)
The number must be even (i.e. it must end in an even digit)

\n" ); document.write( "So there are 5 possible first digits (all but 0).
\n" ); document.write( "There are 6 possible numbers for the 2nd and 3rd digits
\n" ); document.write( "And there are only 3 possible numbers (0, 2 and 4) for the last digit.
\n" ); document.write( "Since the order makes a difference, this is a permutation. We just multiply the number of possible choices for each digit:
\n" ); document.write( "5*6*6*3 = 540 \n" ); document.write( "

\n" );