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Math Help - Ternary sequence

  1. #1
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    Ternary sequence

    A ternary sequence is a sequence of digits made up entirely of 0's, 1's, and 2's such as 002221102. (this is also known as a base three number)

    A. How many 3-digit ternary sequences have exactly one 1?

    B. How many 4-digit ternary sequences have exactly one 1?

    C. How many 4-digit ternary sequences have exactly two 1?

    D. How many n-digit ternary sequences have exactly k 1's (where n and k are non negative ints with k < n?

    For a b and c my answers are

    A. 12
    B. 32
    C. 12

    I did it by hand but was wondering if there is another way to do this with equations? and i do not know how to do D. Can someone explain please.
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  2. #2
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    Hello, smoothi963!

    A ternary sequence is a sequence of digits made up entirely
    of 0's, 1's, and 2's such as 002221102. (this is also known as a base-three number)

    A. How many 3-digit ternary sequences have exactly one 1?
    We have a 3-digit number: ._ _ _

    The "1" can be placed in any of the 3 slots.
    The second digit must not be another 1; there are 2 choices.
    The third digit also has 2 choices.

    Hence, there are: .3 2 .= .12 three-digit numbers with one 1.



    B. How many 4-digit ternary sequences have exactly one 1?
    We have a four-digit number: ._ _ _ _

    The "1" can be placed in any of the 4 slots.
    The second, third and fourth digits have 2 choices each.

    Hence, there are: .4 2 .= .32 four-digit numbers with one 1.



    C. How many 4-digit ternary sequences have exactly two 1's?
    We have a four-digit number: ._ _ _ _

    The two 1's can be placed in C(4,2) = 6 ways.
    The other two digits have 2 choices each.

    Hence, there are: .6 2 .= .24 four-digit numbers with two 1's.



    D. How many n-digit ternary sequences have exactly k 1's?
    (where n and k are non negative integers with k < n)
    Generalize what we did before . . .


    We have an n-digit number: ._ _ _ . . . _

    The k 1's can be placed in: .C(n,k) ways

    Each of the other n-k digits have 2 choices each.
    . . There are: .2^(n-k) ways

    There are: .C(n,k)2^(n-k) n-digit numbers with k 1's.
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  3. #3
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    Thank you
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