P&C - two conditions

Your solution follows the condition that there are NO REPEATED digits, but the question seems to state that only the 9 cannot be repeated.
So, for example, it seems that 2229 is a valid number.

By the way, what are the answer choices? And what's the source of this question?

Cheers,
Brent

Hi student4gmat,

If the "intent" of this question is to create all the 4-digit numbers that are ODD, use the digits 1-9 (but NOT 0) and include the digit 9 just once…..

Since the 9 can be any of the 4 digits, you can break down the number of possibilities into 4 sub-groups (and you'll notice that most of the calculations are exactly the same):

1) The 9 is first: (1)(8)(8)(4) = 256

2) The 9 is second: (8)(1)(8)(4) = 256

3) The 9 is third: (8)(8)(1)(4) = 256

4) The 9 is fourth: (8)(8)(8)(1) = 512

Total possibilities = 1280

If the intent is for 4-digit numbers that are ODD, using the digits 1-9 (but not 0) with one 9, and NO REPEAT DIGITS, then the math changes…

1) The 9 is first: (1)(8)(7)(4) = 224

2) The 9 is second: (8)(1)(7)(4) = 224

3) The 9 is third: (8)(7)(1)(4) = 224

4) The 9 is fourth: (8)(7)(6)(1) = 336

Total possibilities: 1008

GMAT assassins aren't born, they're made,
Rich