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What are the Permutations of the word 'Saskatchewan'?

  1. Oct 10, 2004 #1
    I missed the day when my teacher went over Permutations. If someone could help me with the questions below, that would be great.

    What are the Permutations of the word "Saskatchewan"?

    10 PrN(right?) 6 = The amount of different ways 10 units can be organized into 6 units?

    6! = 6 x 5, 6 x 4 ...... 6 x 1

    8! / 9! = ? What is the purpose of using this and what does it mean?

    Thanks~
     
  2. jcsd
  3. Oct 10, 2004 #2
    You must account for the repetitions of letters in the word.
     
  4. Oct 10, 2004 #3

    HallsofIvy

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    "Saskatchewan" has 12 letters. If they were all different, the answer would be 12!

    However, three of the letters are "a", 2 of the letters are "s" (we don't treat the "S" and "s" as different, do we?) so we could swap the "a"s around without changing the actual word- there are 3! ways to do that. Since we don't want to count those as different, we need to divide by 3! to cancel those. There are 2! ways swap only the "s"s so we also need to divide by 2!: The total number of ways to permute "Saskatchewan" is 12!/(3!2!) (or 12!/3! if the "S" and "s" are considered different.

    No, 6! is not what you say: 6!= 6x 5x 4x 3x 2x 1 = 720.

    8!/9! = 8x7x6x5x4x3x2x1/9x8x7x6x5x4x3x2x1= 1/9 since everything else cancels out.

    I have no idea what your purpose is in using it!
     
  5. Oct 10, 2004 #4
    How many ways can the letters of the following words be arranged?

    Saskatchewan = 39916800
    interesting = 2494800
    Mississippi = 34650

    I need to know how to find those answers. Thanks for your help so far. I'm going to take my Math book upstairs in a bit and stare at it for awhile, that might help.
     
  6. Oct 10, 2004 #5
    The problem is equivalent to arranging P things of which m are alike of one kind, n are alike of another kind, and so on. As mentioned by Sirus earlier, you must account for the repititions in the letters. Once you know how to do this the general way is to plug the values of P, m, n etc into the formula. Since you have trouble understanding how it works, consider the total number of permutations of a word containing x alphabets (of which some may be alike and others distinct) which is x!. If there is at least one alphabet which repeats (say Q) then you will end up with permutations like QQ....or ...QQ... and so on. You can see that the Q's when together can be permuted or arranged in only one way because all of them are identical. But saying that x! is the total number of ways therefore includes these identical arrangements which we must weed out. Thats why you divide by the product of the factorials of the number of different alike things.
     
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