The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,....9
such that each digit does not occur more than once in each number is
(b) 152
(c) 154
(d) None
(a) 150
Answers
Step-by-step explanation:
We need to have a number less than 1000 which means we can have at max three digit number.
Suppose we have the digits as
_ _ _
Case 1: Single Digit Number
0 0 5
Only 1 possible case.
Case 2: Two digit number.
_ 0 -> For this case we can have any digit from 1 to 9 replacing the _ , which means 9 possible cases.
_ 5 -> For this case we can have the digits from 1 to 9 except 5 ( since its already used) , i.e. 8 possible cases.
So total number of cases = 9+8=17
Case 3: Three Digit Number
_ _ 0 -> For this case, we can have the digits from 1 to 9 i.e. 9 different numbers in the tens place and digits from 1 to 9 excluding the digit used in tens place, i.e. 8 different digits in hundreds place. Therefore, 9*8=72 different cases.
_ _ 5 -> For this case, we can have the digits from 0 to 9 (except 5) in tens place, i.e. 8 different digits and any digit excluding 0 and the one used in tens place, i.e. 8 possible digits. Therefore, 8*8=64 different cases.