How many six digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 exactly one such that they are divisible by the digit at the unit's place?
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96*4=384
no ending with 5 and divisible by 5
4*4*3*2*1=96
no ending with 4 and divisible by 4
0
no ending with 3 and divisible by 3
4*4*3*2*1=96(sum of no is always 15,which is divisible by 3)
no ending with 2 and divisible by 2
96
no ending with 1 and divisible by 1
96
total=96*4=384
no ending with 5 and divisible by 5
4*4*3*2*1=96
no ending with 4 and divisible by 4
0
no ending with 3 and divisible by 3
4*4*3*2*1=96(sum of no is always 15,which is divisible by 3)
no ending with 2 and divisible by 2
96
no ending with 1 and divisible by 1
96
total=96*4=384
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