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

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  • MVB
    Lv 6
    9 years ago
    Favorite 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

  • 9 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
  • 9 years ago

    8C2 = 28

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