# How to find the number of $4$-digit odd numbers that can be formed using the digits $3, 4, 5, 6, 7$ without repetition

A $$4$$-digit number will be formed from digits $$3,4,5,6,7$$. Given that digits cannot repeat, find the amount of numbers that can be formed if all numbers are not multiple of $$2$$.

• Hint: The numbers that are not multiples of 2 can (in this case) only end in 3, 5, or 7. That means that the last digit can only be 3, 5 or 7. How would you continue from here? – Matti P. Oct 23 '18 at 6:56
• Welcome to MathSE. When you pose a question here, it is expected that you share your own thoughts on the problem. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Oct 23 '18 at 9:17

A 4 digit number must be formed which is not be a multiple of 2, and so we know that the number must not end with an even number (must not end with 6 or 4 in this case).

We are given 5 digits to work with and we must form a 4 digit number and so 1 of these digits must be omitted in each number formed.

There are two scenarios to form the 4 digits:

scenario 1: 3 odd and 1 even number: If we were to omit out an even number (4 or 6) we are left with 3 odd and 1 even number. For simplicity sake we will take out the 4 first. The even number must not be the last digit, it must be in one of the first 3 digits. There are 6 (3!) ways to arrange the 4 (and 2 odd numbers) in the first 3 digits. we must multiply 6 by 3 as the last digit can be any of the 3 odd numbers so we get 18 ways to rearrange the 4 digit number with 3 odd numbers and 4. multiply 18 by 2 to get total number of ways to rearrange the 4 digit number when we have 3 odd and 1 even number as there are 2 odd numbers, this gives us a total of 36.

scenario 2: 2 odd and 2 even: If we were to omit out an odd number (3,5 or 7) we are left with 2 odd and 2 even number. Again we must have the even numbers exist only in the first 3 digits and there are 6 (3!) ways to rearrange the first 3 digits with 1 odd and 2 even numbers. Multiply by 2 as there are 2 odd numbers in this 4 digit number to get 12 ways to rearrange the 4 digit number. Finally multiply 12 by 3 as any of the 3 odd numbers can be omitted giving you a total of 36 ways to rearrange 2 odd and 2 even.

Total = scenario 1 + scenario 2

Total = 72 number of ways to form an odd 4 digit number from the given 5 digits

A better way:

Mark Bennet offered a much better way of solving it:

The last digit must be either 3,5 or 7. So lets just arbitrarily choose 3 as our last digit, this gives us $$3! \times {4 \choose 3}$$ (choosing 3 digits from the 4 remaining digits), This gives us 24. To get total we must multiply 24 by 3 (as there are 3 digits we can choose to be our last digit) to get a total of 72.

• I would choose the least significant (units) digit first - then you don't need to split into cases - just choose the others from what remains. – Mark Bennet Oct 23 '18 at 7:42