Question

How many 8-digit numbers are there the sum of whose digits is even ?

Answer 1

All are even => (5^8 - 5^7) numbers ------------(1)

Here negative part 5^7 represents numbers with 0 as first digit

Two are odd => 8C2*(5^4)*5^6 - 7C2*(5^2)*5^5 -------------(2

)Four are odd => 8C4*(5^4)*5^4 - 7C4*(5^4)*5^3 -------------(3)

Six are odd => 8C6*(5^6)*5^2 - 7C6*(5^6)*5 ----------------(4)

All are odd => 5^8 -----------(5)

=> Total possibilities = (1) + (2) + (3) + (4) + (5)

= 43906250 ways