How many 5-digit numbers can be formed from the integers 1,2,...,9if no digit can appear more than twice? (For instance, 41434is not allowed.)

Short Answer

Expert verified

5-digit numbers that can be formed are 52920.

Step by step solution

01

Step 1. Given information.

We have to form 5digit numbers using the integers 1,2,...,9.

No digit can appear more than twice.

So, there will be three mutually exclusive even in which this can happen.

02

Step 2. Find the number of ways in which all 5 digits appear only once.

The no. of ways of choosing first digit is 91=9

The no. of ways of choosing first digit is 81=8

The no. of ways of choosing first digit is 71=7

The no. of ways of choosing first digit is 61=6

The no. of ways of choosing first digit is 51=5

Therefore, the number of ways in which all 5 digits appear only once =9×8×7×6×5=15120

03

Step 3. Find the number of ways in which one digit appears twice.

No. of ways in which we can select the 1digit that will appear twice is =91=9

No. of ways in which we can select 2 spots to place the 1 digit that will appear twice is =52=10

The no. of ways of selecting 3digits out of the remaining 8digits is =8×7×6=336

Therefore, the number of ways in which one digit appears twice =9×10×336=30240

04

Step 4. Find the number of ways in which two digits appear twice.

No. of ways in which we can select the 2digits that will appear twice is=92=36

No. of ways in which we can select 4spots to place the 2digits that will appear twice is=54=5

The no. of ways to place the selected 2digits at 4spots such that doubles can be avoided is=4!2!2!=6

The no. of ways of selecting 5th digit out of the remaining 7 digits is =71=7

Therefore, the number of ways in which two digits appear twice=36×5×6×7=7560

05

Step 5. Find the no. of 5-digits numbers that can be formed.

Therefore, the no. of 5-digits numbers that can be formed are=15120+30240+7560=52920

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