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# How many eight-bit strings have exactly two 1's?

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- MVBLv 69 years agoFavorite Answer
just counting it out you'd have:

11000000, 10100000, 10010000... 10000001 for a total of seven with a one in the 1st position;

similarly, six with a one in the 2nd position...

+ 5 + 4 + 3 + 2 + 1 for the first one in the 7th position, for a total of 28.

This is also a combinatorics problems with n=8 and k=2,

C = 8!/6!2! = 56/2 = 28

- Wile E.Lv 79 years ago
00000011

00000101

00001001

00010001

00100001

01000001

10000001

00000110

00001010

00010010

00100010

01000010

10000010

00001100

00010100

00100100

01000100

10000100

00011000

00101000

01001000

10001000

00110000

01010000

10010000

01100000

10100000

11000000

7 + 6 + 5 + 4 + 3 + 2 + 1 = 28

...

Source(s): 12/11/11

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