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

Question

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 $2$.

Answer 1

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.