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Counting Theory and Addition Rule Confuses Me!

  1. Jul 26, 2004 #1
    How many 16-bit strings contain exactly nine 1's? I said the total number is 2 to the 16th power, so to get half with 1's would be 2 to the 8th, or 256.

    How many 16-bit strings contain at least fourteen 1's? I have no idea.

    When determining a how to get 2 pair in a poker hand, why is getting the denomination of the first pair, then the 2 cards, then the demonitation of the second pair, then the 2 cards wrong? I know you have to get the two denominations first, 13 choose 2 instead of 13 choose 1 and then 12 choose 1, but I don't know why.
     
  2. jcsd
  3. Jul 26, 2004 #2

    AKG

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    I have no idea what you did here. The number of 16-bit strings with nine 1's is equal to the number of ways you can choose 9 spots in 16 (to fill them with ones, leaving the rest with zeroes). That number is:

    [tex]{{16}\choose {9}} = 11440[/tex]
    [itex]{{16}\choose {14}} + {{16}\choose {15}} + {{16}\choose {16}} = 137[/itex]*.
    Notice that:

    [tex]{{13}\choose {2}} = \frac{1}{2} \times {{13}\choose {1}} {{12}\choose {1}}[/tex]

    That is, the stuff on the left is 1/2 the stuff on the right. Now why is the stuff on the left right? Because it correctly treats choosing 5 and J as denominations the same as choosing J and 5, whereas your approach treats them as two different cases, so you'd have to divide by 2 (which is already done if you do it the way you know you're supposed to).

    *EDIT: Missed the part that said, "at least"
     
    Last edited: Jul 27, 2004
  4. Jul 27, 2004 #3
    I don't think [itex]{{16}\choose {14}}[/itex] is the correct answer to the second problem, seeing as he's asking for the number of 16-bit strings with /at least/ 14 ones. That means you have to count the number of strings with 15 ones and 16 ones as well. So the answer should be:

    [tex]{{16}\choose {14}} + {{16}\choose {15}} + {{16}\choose {16}}.[/tex]
     
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