*We thoroughly check each answer to a question to provide you with the most correct answers. Found a mistake? Let us know about it through the REPORT button at the bottom of the page.*

How many strings of four decimal digits

a) do not contain the same digit twice?

b) end with an even digit?

c) have exactly three digits that are 9s?

**Contents**hide

## Explanation

**Division rule**

If a finite set A is the union of pairwise disjoint subsets with d elements each, then n = A/d.

**Product rule**

If one event can occur in m ways AND a second event can occur in n ways, the number of ways the two events can occur in the sequence is then m * n.

**Subtraction rule**

If an event can occur either in m ways OR in n ways (overlapping), the number of ways the event can occur is then m + n decreased by the number of ways that the event can occur commonly to the two different ways.

**Sum rule**

If an event can occur either in m ways OR in n ways (non-overlapping), the number of ways the event can occur is then m + n.

## Solution

A) There are 10 possible digits from 0 to 9.

First digit: 10 ways

Second digit: 9 ways (since the digit cannot be the first digit)

Third digit: 8 ways (since the digit cannot be the first nor second digit)

Fourth digit: 7 ways (since the digit cannot be the first, second nor third digit)

Use the product rule: 10 * 9 * 8 * 7 – 5040

B) There are 10 possible digits from 0 to 9.

First digit: 10 ways

Second digit: 10 ways

Third digit: 10 ways

Fourth digit: 5 ways (since there are 5 even digits)

Use the product rule: 10 * 10 * 10 * 5 = 5000

C) There are 10 possible digits from 0 to 9.

First nine: 1 way

Second nine: 1 way

Third nine: 1 way

Other digits: 9 ways (since there are 10 digits and the third digit can’t be 9)

Position 4’s: 4 ways (string is 999x, 99×9, 9×99 or x999 with x digit other than 9)

Use the product rule: 1 * 1 * 1 * 9 * 4 = 36

## Results

A) 5040 strings

B) 5000 strings

C) 36 strings