1. How many 5-digit numbers are there for which the last digit is not a 6 or an 8, the digit 6 appears exactly once, and the number is even

8019 possible 5 digit numbers

Step-by-step explanation:

Available number to pick from are a total of 10 which are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

>> Now, let's consider when the first digit is 6 since we are told that 6 can appear only once.

For the first digit, it's only 1 way.

For the 2nd, 3rd, and 4th digit, we have 9 digits to pick from. Which is 9 ways each.

While for the 5th digit, it can't be 6 or 7. And it has to be even and so, we can pick only 0, 2, 4. Thus, we have 3 ways to choose.

So total number of ways here = 1 × 9 × 9 × 9 × 3 = 3 × 9³ ways

>> Now, let's consider when the 2nd digit is 6 since we are told that 6 can appear only once.

For the 1st digit, since 6 is not possible, and also 0 is not possible because a 5 digit number can't start with zero as it would no longer be a 5 digit number.

Thus, there are 8 ways to choose here.

For the second, third and fourth digits, there will be 9 ways each to chops.

While for the fifth digit, there will be 3 ways to choose.

Total number of ways here = 8 × 1 × 9 × 9 × 3 = 24 × 9² ways

>> When the third digit is 6, number of ways to choose will be same as above = 24 × 9² ways

>> When the fourth digit is 6, number of ways to choose will be same as above = 24 × 9² ways

We can't say none of the digits is 6 because we are told it appears exactly once.

Thus, total possible 5 digit numbers are;

(3 × 9³) + 3(24 × 9²) = 8019 possible 5 digit numbers

b. x- 48 =120

step-by-step explanation:

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