This GMAT permutation combination practice question is a classic counting methods question focusing on numbers and number properties. You can use the explanation given in this question as a template to solve questions of this kind. After cracking the question, create your own variants of this question and try solving them to master this idea.
How many odd 4 digit positive integers that are multiples of 5 can be formed without using the digit 3?
Choice D. 648
Let us breakdown what is stated in the question into bite sized pieces. The numbers that we are counting in this question should satisfy the following 4 conditions.
- The number is a 4 – digit number.
- It is an odd number.
- It is a multiple of 5.
- 3 cannot be a digit in the number.
Step 1: Let us start with the units place – the right most digit of the number
Any number that is divisible by 5 will end in 0 or 5. Combining this information with condition 2 that the number is an odd number, we can conclude that the unit digit of all these numbers is 5.
Therefore, the unit digit has only one option. That was easy. One out of the 4 digits out of our way.
Step 2: Let us find out the number of options for the left most digit
The thousands place, the left most digit of these numbers cannot be 0.
The question stipulates that 3 cannot be a digit of the number. So, it cannot be 3.
Out of the total 10 possibilities, 0 to 9, we cannot use 2 digits.
That leaves us with 8 options for the thousands or the left most digit.
Step 3: The remaining two digits – hundreds and tens place
a. The hundreds place cannot be 3. It can take any value from 0 to 9 except 3. So, it has 9 options.
b. The tens place also cannot be 3. Using the same argument that we used for hundreds place, the tens place of these numbers has 9 options.
Therefore, the number of 4 digit numbers that can be formed that satisfy all of the mentioned conditions
= 8 * 9 * 9 * 1 = 648 numbers.
elok darojatin says
good posting sir,
may i know your e-mail please?