Results 1 to 5 of 5

Math Help - Combinations Help

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    25

    Combinations Help

    This is a large word problem. I have stared at it for 2 days and I just can't wrap my mind around where to start. Any help in the right direction would be greatly appreciated!

    Suppose that for a given computer, a password must satisfy each of the following conditions:

    1) The password must be between 2 and 6 characters
    2) The first character must be a letter and the last character must be a digit between 0 & 9.
    3) for passwords with 3 or more characters, all characters except for the first or last may be either a letter or a digit.
    4) repetition is allowed.
    5) they are not case sensitive.

    How many passwords are possible for this computer?

    I did it like this:

    (26C4)+(36C3)+(36C2)+(10C1)=26300. This does not look right to me!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,862
    Thanks
    742
    Hello, ezwind72!

    These are not combinations!
    With a password, the order of the characters is important.


    Suppose that a password must satisfy the following conditions:

    1) The password must be between 2 and 6 characters
    2) The first character must be a letter and the last character must be a digit (0 - 9).
    3) All characters except for the first or last may be either a letter or a digit.
    4) Repetition is allowed.
    5) They are not case sensitive.

    How many passwords are possible?

    The first must be a letter: 26 choices.
    The last must be a digit: 10 choices.
    The others must be a letter or a digit: 36 choices.

    \begin{array}{c|c} \text{Length} & \text{Choices} \\ \hline<br />
2 & 26\cdot10 \\ 3 & 26\cdot36\cdot10 \\ 4 & 26\cdot36\cdot36\cdot10 <br />
\\ 5 & 26\cdot36\cdot36\cdot36\cdot10 \\ <br />
6 & 26\cdot36\cdot36\cdot36\cdot36\cdot10 \end{array}


    Total: . 260 + 260\cdot36 + 260\cdot36^2 + 260\cdot36^3 + 260\cdot36^4

    . . = \;260\left(1 + 36 + 36^2 + 36^3 + 36^4\right) \;=\;260\,\frac{36^5-1}{36-1}

    . . =\;260\,(1,\!727,\!605) \;=\;\boxed{449,\!177,\!300}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    25
    Quote Originally Posted by Soroban View Post
    Hello, ezwind72!

    These are not combinations!
    With a password, the order of the characters is important.



    The first must be a letter: 26 choices.
    The last must be a digit: 10 choices.
    The others must be a letter or a digit: 36 choices.

    \begin{array}{c|c} \text{Length} & \text{Choices} \\ \hline<br />
2 & 26\cdot10 \\ 3 & 26\cdot36\cdot10 \\ 4 & 26\cdot36\cdot36\cdot10 <br />
\\ 5 & 26\cdot36\cdot36\cdot36\cdot10 \\ <br />
6 & 26\cdot36\cdot36\cdot36\cdot36\cdot10 \end{array}


    Total: . 260 + 260\cdot36 + 260\cdot36^2 + 260\cdot36^3 + 260\cdot36^4

    . . = \;260\left(1 + 36 + 36^2 + 36^3 + 36^4\right) \;=\;260\,\frac{36^5-1}{36-1}

    . . =\;260\,(1,\!727,\!605) \;=\;\boxed{449,\!177,\!300}

    I understand where you are coming from with the first being 26 choices and the last being 10. I am confused why it would be 26x10 though. I just loathe these word problems! I appreciate all your help though.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2008
    Posts
    25
    BTW, you said it isn't a combination. Does that make it a permutation?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member Rheanna's Avatar
    Joined
    Nov 2008
    Posts
    36
    Awards
    1
    ooh hope I get this one as an extra credit ;p
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Combinations in a set
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: September 9th 2010, 07:19 AM
  2. How many combinations are possible?
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: July 23rd 2009, 08:53 PM
  3. Combinations
    Posted in the Statistics Forum
    Replies: 2
    Last Post: May 5th 2008, 09:28 AM
  4. How many combinations..?
    Posted in the Algebra Forum
    Replies: 9
    Last Post: May 2nd 2008, 11:34 AM
  5. combinations
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 27th 2008, 10:00 AM

Search Tags


/mathhelpforum @mathhelpforum