Question

The total number of $8$ digited numbers which have different digits is:

• A. $8!$
• B. ${}^{10}{p}_g$
• C. ${}^{10}{p}_7$
• D. ${}^{10}{p}_6$

Answer 1

The correct option is A $8!$

Total number or $8$ digit numbers which have different digits is $\left(8\times 7!\right)$

Number of ways filling 1st place =$8$

Number of ways filling remaining place= $7$

$8\times 7\times 6\times 5\times 4\times 3\times 2\times 1=8!$

then the total numbers$=8!$