# Questionbank: Permutation and Probability

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## Counting 5 Digit Numbers

Q.11: How many numbers of up to 5 digits can be created using the digits 1, 2, 3 and 5 each at least once such that they are a multiple of 15?
1. 24
2. 18
3. 15
4. 12

Choice D. 12

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## Detailed Solution

For a number to be a multiple of 15, it has to be a multiple of 3 and of 5. So, the last digit has to be 5 and the sum of digits should be a multiple of 3.

We can have either 4–digit or 5–digit numbers. If we have a 4–digit number, sum of the digits will be 1 + 2 + 3 + 5 = 11. No 4–digit number formed with digits 1, 2, 3, 5 exactly once can be a multiple of 3. So, there is no possible 4–digit number.

Now, in any 5 digit number, we will have 1, 2, 3, 5 once and one of these 4 digits repeating once. 1 + 2 + 3 + 5 = 11. So, the digit that repeats in order for the number to be a multiple of 3 has to be 1. In this instance, sum of the digits will be 12 and this is the only possibility.

So, any 5–digit number has to have the digits 1, 1, 2, 3, 5. For the number to be a multiple of 5, it has to end in 5.

So, number should be of the form __ __ __ __ 5, with the first 4 slots taken up by 1, 1, 2, 3. These can be rearranged in ${4! \over 2!}$ = 12 ways.

There are 12 possibilities overall.

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