Case 1: When the last digit is 9.

First 3 digits = 634/674 or, \(2\) ways

Last digit = 9, or, 1 way

No of ways to fill the remaining 3 digits =\( 9\) ways

Total no of ways when the last digit is 9 = \(2 (for \ 1_{st} \ 3 \ digit) * 9(4^{th}) * 9(5^{th}) * 9(6^{th}) * 1 (Last \ digit) = 729*2\)

Case 2: When 9 in any other position except for the last position, then it can take 3 position and the last position can be {1,3,5,7}

Therefore no of ways \(= 2 (for \ 1_{st} \ 3 \ digit) * 9 * 9 * 3(When \ the \ position \ of \ 9 \ changes) * 4 (Last \ digit) = 972*2\)


Hence the total ways \(= 1458 + 1944 =3402\)

Ans C