A 4 digit number having all distinct digits is such that the product of all digits is 8 less than the sum of all digits. How many such 4 digits number are possible ?
4,3,1,0 in 18 ways & 5,2,1,0 in 18 ways As 0 should not be first digit because it will become 3 digit no.
So 18+18=36 ways.
Hence 36 4-digit no's will be possible.
All the digits of the given number are different.
2 two digit numbers chosen in such a way that all 4 digits used are distinct, are such that their sum is 90. How many such pairs are possible if in both numbers units place is 1 more than tens place?
