Now to tackle the above problem . I am listing below 2 methods below I used to solve it and I got the correct answer using the latter one :-
Restrictions given are:-No digit is repeated
- Number is divisible by 5
Method 1:-
Total number of 3 digit numbers using restriction 1 can be formed as follows
$1^{st}$ digit | $2^{nd}$ digit | $3^{rd}$ digit |
---|---|---|
all digits except 0 are possible |
restriction 1 | restriction 1 |
9 cases | 9 cases , 1 taken already | 8 cases , 2 taken already |
total cases = 9 * 9 * 8 = 648
Total number of 3 digit numbers using restriction 1 but not divisible by 5 can be formed as follows
$1^{st}$ digit | $2^{nd}$ digit | $3^{rd}$ digit |
---|---|---|
all digits except 0 are possible (chosen after $2^{nd}$ and $3^{rd}$ digit) |
after selecting third place use restriction 1 | 0 and 5 not possible |
8 cases | 9 cases , 1 taken already | 8 cases |
total cases = 8 * 9 * 8 = 576
Method 2:-
We make 2 cases for all required 3 digit numbers:- (Pertaining to restrictions above)
- Number ends with 0 = 9*8=72
- Number ends with 5 = 8*8=64