How many odd three-digit integers greater than 800 are there such that all their digits are different?
A. 40
B. 56
C. 72
D. 81
E. 104
A. 40
B. 56
C. 72
D. 81
E. 104
Bia
My answer is 72 . Here is explaination.
Odd numbers between 800 and 900
Start : 801
End : 899
801-809 : 5
811-819 : 4 (For odd tens digit we will have one lesser number like 811 has reperion then 833 will be same etc)
821-829 : 5
831-839 : 4
841-849 : 5
851-859 : 4
861-869 : 5
871-879 : 4
881-889 : 0 (All 88_)
891-899 : 4
Total numbers between 801-899 : 5*4 + 4* 5 = 40
Odd numbers between 900 and 1000
Start : 901
End : 999
901-909 : 4 (as In hundreds place is 9 and all odd number from 1-10 will have atleast one 9 this case 909)
911-919 : 3 (Here we need to account for two one coz of tens place as odd number and other for number ending with 9
two numbers elimnated are 911 and 919)
921-929 : 4
931-939 : 3
941-949 : 4
951-959 : 3
961-969 : 4
971-979 : 3
981-989 : 4
991-999 : 0
Total numbers between 901-999 : 4*5 + 3* 4 = 32
Total Number = 40+32 = 72
