An office manager must choose a five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of the digits is allowed. How many different lock codes are possible.?  Options
A) 3120  B) 3250  C) 3275  D) 3860  E) 3980
However, I think 6720 (the answer given in MGMAT) is wrong. Here's my reasoning. If you have (an) odd number(s) for your 2nd, 3rd and/or 4th digits, you cannot repeat that(those) numbers for the last digit. Therefore, this problem needs to be partitioned into 4 groups.   1) ODD (1st digit); 3 EVENs (2nd, 3rd, 4th digits); ODD (5th digit)   5 x 5x 4 x 3 x 4 = 1200   2) ODD (1st digit); 2 EVEN, 1 ODD (2nd, 3rd, 4th digits); ODD (5th digit)   5 x 5 x 4 x 4 x 3 = 1200   3) ODD (1st digit); 1 EVEN, 2 ODDs (2nd, 3rd, 4th digits); ODD (5th digit)   5 x 5x 4 x 3 x 2 = 600   4) ODD (1st digit); 3 ODDs (2nd, 3rd, 4th digits); ODD (5th digit)   5 x 4 x 3 x 2 x 1 = 120   Then the total # of combinations would be 1200 + 1200 + 600 + 120= 
<gadget id="calculator">1200 + 1200 + 600 + 120</gadget>
<output>3_120</output>
3120.   3120NS:3120
<result>A</result>
An office manager must choose a five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of the digits is allowed. How many different lock codes are possible.?