How many total numbers of seven-digit numbers can be formed having sum of whose digits is even is
How many total numbers of seven-digit numbers can be formed having sum of whose digits is even is
(a) 9000000 (b) 4500000 (c) 8100000 (d) 4400000 (e) None of these
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(b) Suppose x1 x2 x3 x4 x5 x6 x7 represents a seven digit
number. Then x1 takes the value 1, 2, 3, ....., 9 and x2, x3,
....., x7 all take values 0, 1, 2, 3, ......, 9.
If we keep x1, x2, ......, x6 fixed, then the sum x1 + x2 + ......
+ x6 is either even or odd. Since x7 takes 10 values 0, 1,
2, ....., 9, five of the numbers so formed will have sum of
digits even and 5 have sum odd.
Hence the required number of numbers
= 9 . 10 . 10 . 10 . 10 . 10 . 5 = 4500000.
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