Q. 234.7( 3 Votes )

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

Answer :

Given: Nine-digit number is required in which the first digit cannot be zero and the repetition of digits is not allowed.



Assume nine boxes, now the first box can be filled with one of the nine available digits, so the possibility is 9C1


Similarly, the second box can be filled with one of the nine available digits, so the possibility is 9C1


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


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


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


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


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


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


the ninth box 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!)


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