Total number of eight digit numbers in which all digits are different is

Question

Total number of eight digit numbers in which all digits are different is (A) (8.7 !/3) (B) (5.8 !/3) (C) (8.9 !/2) (D) (9.9 !/2)

Tardigrade Admin · Accepted Answer

There are ten digits 0,1,2, ldots ldots ldots .9 . Permutations of these digits taken eight at a time = 10 P 8 which includes permutations having 0 at the first. When 0 is fixed at the first place, then number of such permutations = 9 P 7 So, required number = 10 P 8- 9 P 7 =(10 !/2)-(9 !/2)=(9.9 !/2)

Q. Total number of eight digit numbers in which all digits are different is

$\frac{8.7 !}{3}$

$\frac{5.8 !}{3}$

$\frac{8.9 !}{2}$

$\frac{9.9 !}{2}$

Q. Total number of eight digit numbers in which all digits are different is

38.7!​

35.8!​

28.9!​

29.9!​

There are ten digits 0,1,2,……….9. Permutations of these digits taken eight at a time =10P8​ which includes permutations having 0 at the first. When 0 is fixed at the first place, then number of such permutations = 9P7​ So, required number =10P8​−9P7​ =210!​−29!​=29.9!​

$\frac{8.7 !}{3}$

$\frac{5.8 !}{3}$

$\frac{8.9 !}{2}$

$\frac{9.9 !}{2}$