gmattesttaker2 wrote: How many odd positive integers n are there such that 40,000 < n < 100,000 and n is made up of the digits 0, 1, 2, 3, 4 and 5?

A) 1296
B) 1295
C) 750
D) 749
E) 150

OA: A

I was trying to solve as follows:

The number of digits are 5.

The last digit could be 1 or 3 or 5. Hence, we have 3 choices for the last digit.

The first digit has to be 4 or 5. Hence, we have 2 choices for the first digit.

Now, the digit could be say 41,111 or 41,113 or 41,114 or 41,235

Hence, the number of integers are: ( 2 x 6 x 6 x 6 x 3 )/ ( 3!)

I am dividing by 3! to avoid double counting the repeating digit in the 2nd, 3rd and 4th place.

GMATGuruNY wrote:

gmattesttaker2 wrote: How many odd positive integers n are there such that 40,000 < n < 100,000 and n is made up of the digits 0, 1, 2, 3, 4 and 5?

A) 1296
B) 1295
C) 750
D) 749
E) 150

OA: A

Number of options for the ten-thousands digit = 2. (4 or 5.)
Number of options for the thousands digit = 6. (0, 1, 2, 3, 4, or 5.)
Number of options for the hundreds digit = 6. (0, 1, 2, 3, 4, or 5.)
Number of options for the tens digit = 6. (0, 1, 2, 3, 4, or 5.)
Number of options for the units digit = 3. (1, 3, or 5.)
To combine these options, we multiply:
2*6*6*6*3 = 1296.

The correct answer is A.

I was trying to solve as follows:

The number of digits are 5.

The last digit could be 1 or 3 or 5. Hence, we have 3 choices for the last digit.

The first digit has to be 4 or 5. Hence, we have 2 choices for the first digit.

Now, the digit could be say 41,111 or 41,113 or 41,114 or 41,235

Hence, the number of integers are: ( 2 x 6 x 6 x 6 x 3 )/ ( 3!)

I am dividing by 3! to avoid double counting the repeating digit in the 2nd, 3rd and 4th place.

Your solution is perfect until the portion in red.
We would divide by 3! if the order of the 3 middle digits didn't matter -- if 41235 were considered the SAME result at 42315.
But 41235 is a DIFFERENT integer from 42315.
Thus, there is no reason to divide by 3!.