Probability and Permutations

Add yours
Introduction to Permutations and Combinations
8:31 — by statisticsfun

Tip: This isn't the place to ask a question because the teacher can't reply.

Key Questions

  • Permutations of items are arrangements of items.


    Examples

    All six permutations of #{a,b,c}# are:

    #{abc, acb, bac, bca, cab, cba}#

    Also, all 6 permutations of #{a,b,c}# chosen at 2 items at a time are

    #{ab, ba, ac, ca, bc, cb}#


    I hope that this was helpful.

  • Permutations are arrangements of items, so the number of permutations is the number of arrangements of items.

    Let #P(n,r)# denote the number of permutations of #n# items chosen #r# items at a time. #P(n,r)# can be found by

    #P(n,r)=n cdot (n-1) cdot (n-2)cdot cdots cdot(n-r+1)={n!}/{(n-r)!}#.


    Example

    How many 3-digit codes are possible if each digit is chosen from 0 through 9, and no digits are repeated.

    We can think of 3-digits codes as permutations of #10# digits chosen #3# digits at a time since no digits are repeated. So, we have

    #P(10,3)=10 cdot 9 cdot 8=720#.

    Hence, there are 720 possible 3-digit codes.


    I hope that this was helpful.

  • This key question hasn't been answered yet. Answer question
  • I am assuming that you are using a Ti-83+ or Ti-84. The following is a direction for these calculators. However, similar menus can be found on just about any calculator.

    Permutations is a method of counting when order matters. The general form is #""_nP_r#, where #n# is the total number of choices, and #r# is the total number of choices. The formula for #""_nP_r = (n!)/((n-r)!)#

    Try this for a problem: There is a race of 20 people. In how many ways can there be a first, second, and third? Note, order does matter! Coming in first is different than coming in second. This is why we use Permutations. Now, know your variables. The total number of people there are is 20, which is your #n#. You want the top three, so that's your #r#. Thus, #n = 20# and #r = 3#, giving you #""_(20)P_3#.

    Method 1: the long way.

    You can use the formula given above. Substitute in your values, then entering the values in your calculator. So, #""_(20)P_3 = (20!)/((20-3)!) = (20!)/(17!) = (20*19*18*17!)/(17!) = 20*19*18 = 6,840#.

    This can get tedious, especially if you can't remember the formula.

    Method 2: the better way!

    Most calculators have the formulas stored. To access these formulas, follow these steps.

    • Turn the calculator on, go to home screen (the main screen to do the calculations).
    • Write in the #n#. In our example, enter in 20.
    • Go into the MATH menu by pressing the MATH key.
    • Move to the right, entering into the Prob menu.
    • Find the #""_nP_r# selection and select it.
    • Write in the #r#. In our example, enter in 3.
    • Your screen should look like: 20 #""_nP_r# 3; hit ENTER.
    • The answer: 6,840 should show up.

    Most calculators have a Prob button somewhere. They are done in the similar way as Method 2.

    Happy Mathing!

  • It is important in permutations by the definition of the word "permutation". If we ignore order, then we are discussing / choosing "combinations".

    Given set #{a, b, c, d}#, if we count #ab# and #ba# as the same, then we are counting combinations.
    If they are counted as different, then we are counting permutation.

Questions

  • Ken C. answered · 5 months ago