GATE Overflow Test Series | Discrete Mathematics | Test 2 | Question: 7

Question

How many $10$ digit numbers greater than $1987654321$ have no two digits same?

• A. $8 \times 9!$
• B. $9 \times 9!$
• C. $7 \times 9!$
• D. None of the above

Answer 1

The next integer number after $1987654321$ without a digit being repeated must start with $2.$ So, our problem is same as the number of $10$ digit integers having no repeated digits and not starting with $1.$

Now, we have $8$ possibilities for the first digits (it cannot start from $0$ or $1$). For the second digit we have $9$ possibilities (excluding the first selected one), for the third digit $8$ possibilities and so on we get $1$ possibility for the last digit.
So, total possibilities $ = 8 \times 9!$

Comments

Can anyone explain this in detail?

Not able to understand the given solution.

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Here you can see that the number given has 10 digits and 1 is occurring two times. We need to count the numbers of 10 digits which are greater than the given number 1987654321 without any digit occurring more than once. So we have to use all 10 digits (0-9) once. For numbers to be greater than 1987654321, we can start from (2-9) as first digit. (First digit can not be 0 or 1 as it violates the given condition being greater than 1987654321). So we have 8 choices for 1st place. Now out of remaining 9 digits(now we can use 0,1), we can put them in 9! ways at remaining 9 places. So total 8 * 9! ways.

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this is very good question!