A three-digit number is formed from the digits 0,1,3,5,6,7 and 8. Each digit can be used by once. How many of these are even numbers?

n = 80

Step-by-step explanation:

The number of digits that can be used are 7, which are 0,1,3,5,6,7,8. Now we have to form a 3 digit number that is even. So, for a number to be even, its last digit must also be an even digit or a zero.

FOR LAST DIGIT TO BE NON-ZERO

In this case we will consider all the numbers that have a last non-zero digit. First we choose the no. of options for the last place, that is unit place. There will be 2 options for unit place (6,8). Because only these digits will result in an even number

No. Of Options For Unit Place = 2

Now, we see number of options for hundredth place. Since, one of the digits 6 and 8 is already used and can not be used again and the digit 0 can also not be used on hundredth place. Hence,

No. Of Options For Hundredth Place = 7 - 2 = 5

Now, for tenth place we will subtract the 2 options from total that are being used at hundredth and unit place:

No. Of Options For Tenth Place = 7 - 2 = 5

So, the no. of ways to to form a three digit even number with a non-zero last digit (n₁) are:

n₁ = (2)(5)(5)

n₁ = 50

FOR LAST DIGIT TO BE ZERO

In this case we will consider all the numbers that have a last digit to be zero. So, there is only one option for the unit place, that is 0.

No. Of Options For Unit Place = 1

Now, we see number of options for hundredth place. Since, 0 is already used at unit place, so the rest of the digits can all be used at hundredth place.

No. Of Options For Hundredth Place = 7 - 1 = 6

Now, for tenth place we will subtract the 2 options from total that are being used at hundredth and unit place:

No. Of Options For Tenth Place = 7 - 2 = 5

So, the no. of ways to to form a three digit even number with zero as last digit (n₂) are:

n₂ = (1)(6)(5)

n₂ = 30

Now, for the total no. of ways to form a 3 digit even number from the given numbers without repetition (n) are:

n = n₁ + n₂

n = 50 + 30

n = 80