How many odd three-digit integers greater than 800 are there

[BTGmoderatorLU]

How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 56
C. 72
D. 81
E. 104

The OA is C.

Please, can anyone assist me with this PS question? Thanks in advance!

[regor60]

How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 56
C. 72
D. 81
E. 104

The OA is C.

Please, can anyone assist me with this PS question? Thanks in advance!

So the range is 801 to 999 inclusive. Start with the numbers in the 800's.

First digit is 8. Assume an even number as the second digit. Since 8 is taken there are 4 choices. Third digit has 5 choices. Total choices 1x4x5 = 20

Same thing but assume odd number as second digit. 5 choices. Third digit also odd, but now 4 choices since used one for the second digit. Total choices 1x5x4 = 20.

Now do the 900's in the same way. First digit 9. Second digit even, 5 choices. Third digit odd, 4 choices since 9 is taken, total choices 1x5x4 = 20.

Second digit odd. 4 choices since 9 taken. Third digit odd, 3 choices since two have been taken, total 1x4x3 = 12.

Adding the choices 20 + 20 + 20 + 12 = 72, C

[GMATGuruNY]

How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 56
C. 72
D. 81
E. 104

Case 1: Hundreds digit = 8
Number of options for the units digit = 5. (Any of the 5 odd digits.)
Number of options for the tens digit = 8. (Of the 10 digits, any but the two already used.)
To combine the options above, we multiply:
5*8 = 40.

Case 1: Hundreds digit = 9
Number of options for the units digit = 4. (Of the 5 odd digits, any but 9.)
Number of options for the tens digit = 8. (Of the 10 digits, any but the two already used.)
To combine the options above, we multiply:
4*8 = 32.

Total number of viable integers = Case 1 + Case 2 = 40 + 32 = 72.

The correct answer is C.

[Brent@GMATPrepNow]

Here's a similar one to practice with: https://www.beatthegmat.com/counting-a-3 ... 88550.html

Cheers,
Brent

[Jeff@TargetTestPrep]

How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 56
C. 72
D. 81
E. 104

Since the integer is odd, the last (or units) digit has five choices: 1, 3, 5, 7 or 9.

If the first (or hundreds) digit is 8, then the second (or tens) digit has 8 choices (since that digit has to be different from the hundreds and from the units digit). Thus, there are 1 x 8 x 5 = 40 odd integers in the 800s.

If the first (or hundreds) digit is 9, then the units digit has only 4 choices (since it can't be 9), and the second (or tens) digit has 8 choices (since that digit has to be different from the hundreds and from the units digit). Thus, there are 1 x 8 x 4 = 32 odd integers in the 900s.

Therefore, we have a total of 40 + 32 = 72 such integers.

Answer: C