How many 9-digit numbers of different digits can be formed?

Asked by Aaryan | 1 year ago |  35

##### Solution :-

In a nine-digit number, 0 cannot appear in the first digit and the repetition of digits is not allowed. So, the number of ways of filling up the first digit is 9C1= 9

Now, 9 digits are left including 0. So, the second digit can be filled with any of the remaining 9 digits in 9 ways.

Similarly, the third box can be filled with one of the eight available digits, so the possibility is 8C1

The fourth digit can be filled with one of the seven available digits, so the possibility is 7C1

The fifth digit can be filled with one of the six available digits, so the possibility is 6C1

The sixth digit can be filled with one of the six available digits, so the possibility is 5C1

The seventh digit can be filled with one of the six available digits, so the possibility is 4C1

The eighth digit can be filled with one of the six available digits, so the possibility is 3C1

The ninth digit can be filled with one of the six available digits, so the possibility is 2 C1

Hence the number of total possible outcomes is 9C1 × 9C1 × 8C1 × 7C1 × 6C1 = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 9 (9!)

Answered by Sakshi | 1 year ago

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