Number of odd integers

[psm12se]

How many three-digit odd integers less than 500 are there so that all the digits are not distinct?

A. 56

B. 88

C. 144

D. 200

E. 250

[Anju@Gurome]

How many three-digit odd integers less than 500 are there so that all the digits are not distinct?

While attacking counting problem always satisfy the restrictions first, so that you can freely select the rest. Here the restrictions are,

• 1. Integers must be less than 500 ---> First digit can be either 1, 2, 3, or 4.
  2. Integers must be odd ---> Last digit can be either 1, 3, 5, 7, or 9.

Now, there are (500 - 100)/2 = 200 three-digit odd integers less than 500.
If we can calculate the number of three-digit odd integers less than 500 with all the digits different and subtract that from 200, we will have our answer.

Number of three-digit odd integers less than 500 with all the digits different...

• Numbers starting with 1 ---> (4 choice for last digit)*(8 choice for middle digit) --> 32
  Numbers starting with 2 ---> (5 choice for last digit)*(8 choice for middle digit) = 40
  Numbers starting with 3 ---> Same as numbers starting with 1 --> 32
  Numbers starting with 4 ---> Same as numbers starting with 2 --> 40

Hence, number of three-digit odd integers less than 500 are there so that all the digits are not distinct = 200 - (2*32 + 2*40) = 200 - 144 = 56

The correct answer is A.

[GMATGuruNY]

How many three-digit odd integers less than 500 are there so that all the digits are not distinct?

A. 56

B. 88

C. 144

D. 200

E. 250

Integers with all 3 digits NOT distinct = total number of integers - integers with all 3 digits distinct.

Total number of integers:
To count consecutive integers:
Number of integers = biggest - smallest + 1.
Thus, from 100 to 499, inclusive:
Number of integers = 499 - 100 + 1 = 400.

Integers with all 3 digits distinct:
Number of options for the hundreds digit = 4. (1, 2, 3 or 4).
Number of options for the tens digit = 9. (Any digit 0-9 but the digit already used.)
Number of options for the units digit = 8. (Any digit 0-9 but the 2 digits already used.)
To combine these options, we multiply:
4*9*8 = 288.

Thus, from 100 to 499, inclusive:
Number of integers with all 3 digits NOT distinct = 400-288 = 112.

Of these 112 integers, exactly HALF will be odd:
112/2 = 56.

The correct answer is A.