What are the Permutations of the word 'Saskatchewan'?

  • Thread starter Thread starter Dooga Blackrazor
  • Start date Start date
  • Tags Tags
    Permutations
AI Thread Summary
The discussion focuses on calculating the permutations of the word "Saskatchewan," which consists of 12 letters, including repetitions of the letters "a" and "s." The correct formula for the total permutations is 12!/(3!2!), accounting for the indistinguishable letters. Clarifications are provided regarding the calculations of factorials, specifically that 6! equals 720 and 8!/9! simplifies to 1/9. Examples of permutations for other words like "interesting" and "Mississippi" are also given, emphasizing the need to consider letter repetitions. Understanding this concept is crucial for accurately determining the number of arrangements for any set of letters.
Dooga Blackrazor
Messages
258
Reaction score
0
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~
 
Physics news on Phys.org
You must account for the repetitions of letters in the word.
 
"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!
 
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.
 
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top