How many five-digit even numbers, that are less than 5 and that are not divisible by 5, can be created from the digits 0, 1, 3, 5, 6, 7, 8 and 9, if no digit can be used more than once?

480 numbers

Step-by-step explanation:

We have to create a 5-digit even number less than 50,000 using the digits: 0, 1, 3, 5, 6, 7, 8 and 9

Total available digits = 8

No digit can be used more than once. We have to fill the 5 places using the 8 numbers.

Since, the number must be less than 50,000, the first digit cannot be 5 or greater than 5 and it cannot be 0 either. So there are only two possible choices for the 1st place i.e. either 1 or 3.

2 possible ways , __ , __ , __ , __

Since the digit must be even and cannot be divisible by 5, so the last digit must be event but not 0. So there are 2 ways to fill the last place i.e. using 6 or 8.

2 possible ways , __ , __ , __ , 2 possible ways

When a digit is chosen for 1st and the last place, we will be left with 6 digits which can be chosen in any order. So, the 2nd place digit can be filled in 6 ways. After choosing this digit we are left with 5 digits.

Number of ways to fill the 3rd place = 5

Number of ways to fill the 4th place = 4

2 possible ways , 6 Possible ways , 5 Possible Ways , 4 Possible Ways , 2 possible ways

According to the fundamental counting principle, the number of total possible 5-digit numbers will be equal to the product of possibilities of each individual position.

i.e.

Total number of possible 5-digit numbers = 2 x 6 x 5 x 4 x 2 = 480

Therefore, 480 five-digit numbers can be formed using the given conditions.

can you put the picture up?

step-by-step explanation:

2425

step-by-step explanation: