Question 1205238: How many different arrangements can be formed by taking 4 letters from the word
STATISTICS? Assume that the letters are taken without replacement.
Found 4 solutions by MathLover1, Edwin McCravy, math_tutor2020, greenestamps:
Answer by MathLover1(20850)   (Show Source):
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website!
The above solution is incorrect. She took it to be choosing all 10 letters.
We are only choosing 4 of them.
We will have letters to rearrange of the following forms:
XYZW, XXYZ, XXYY, or XXXY, where the letters X,Y,Z,W can represent
distinguishable letters of STATISTICS. So there are 4 cases.
Case XYZW. The letters are all distinguishable. We choose any 4 from
the 5 letters STAIC C(5,4)=5 ways and arrange them 5!=120 ways.
Case XXYZ. 2 letters are indistinguishable. We choose the letter for
X from the letters STI C(3,2)=3 ways and the Y and Z from the 4 letters,
made up of the 2 letters we didn't choose for X, A and C, which is C(4,2)=6
That's (3)(6)=18 ways, and then we can rearrange the letters  ways.
That's (18)(12)=216 ways.
Case XXYY. The 2 letters for X are indistinguishable. The 2 letters for Y's
are indistinguishable. We choose the letters X and Y from the letters STI
C(3,2)=3 ways. Then we can rearrange the letters  ways.
That's (3)(6)=18 ways.
Case XXXY. We can choose the S or the T for X. That's 2 ways. That leaves 4
ways to choose the letter for Y. That's (2)(4)=8 ways. Then we can rearrange
the letters  ways. That's (8)(4)=32 ways.
Total = 120+216+18+32 = 386 ways.
Edwin
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answer: 386
Explanation
There are 10 letters in the word STATISTICS.
We wish to create 4 letter "words" from this pool of letters.
If we could tell the repeated letters apart with color-coding (eg: STAT vs STAT) or using subscripts (eg: ST1AT2 vs ST2AT1), then the answer would be simply  according to the nPr permutation formula.
The exclamation mark refers to factorials.
However, we don't have the luxury of color-coding nor subscripts.
We cannot tell the repeated letters apart.
We'll have to break things up into disjoint cases.
Case (i): 3 letters are the same, 1 is different (eg: SSST).
Case (ii): 2 letters are the same, 2 are different (eg: SSTA).
Case (iii): 2 sets of twin letters (eg: SSTT).
Case (iv): all 4 letters are different (eg: STAC).
Only one of those cases can happen which is why they're considered disjoint.
There is no overlap between any two cases.
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Case (i)
Let's make a frequency chart
| Letter | Frequency |
|---|
| S | 3 | | T | 3 | | A | 1 | | I | 2 | | C | 1 | | Total | 10 |
The only letters that show up 3 times are S and T.
If you select 3 "S" letters, then the other choices are T, A, I, or C.
Let letter M represent a single letter from the set {T, A, I, C}
Consider the permutations of SSSM.
There are 4 ways to rearrange the letters of SSSM because we have 4 places to put the M.
Once the dust settles, we need to replace M with T, A, I, or C. There are 4 choices to pick here.
So we have 4*4 = 16 permutations where three "S" letters are chosen.
Through similar logic, there are also 16 permutations of the form TTTN, where N is a placeholder for a single letter of the set {S, A, I, C}.
There are 16+16 = 32 ways to do case (i).
Let x = 32 since we'll use it later.
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Case (ii)
Refer to the frequency chart made in case (i).
Here are the letters that show up at least twice: S, T, and I
There are 3 ways to pick the twin letters.
Use the nCr combination formula to find there are 4C2 = 6 ways to pick the two distinct letters from a pool of four.
I'll leave the scratch work steps for the student to do.
Let's say you picked S as the twin letters and T,A as the other two distinct letters.
So we're trying to permutate SSTA.
There are (4!)/(2!) = 12 ways to rearrange the letters of SSTA. I divided by 2! to correct for double-counting.
Recap
3 ways to pick the twin letters
6 ways to pick the two distinct letters
12 ways to permute those four letters chosen
Overall there are 3*6*12 = 216 ways to carry out case (ii).
Let y = 216 so we can use it later.
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Case (iii)
Refer to the frequency chart made in case (i).
Once again, the letters that show up at least twice: S, T, and I
There are 3 items listed. There are 3 ways to pick two of those letters (i.e. 3 ways to leave a certain letter out).
Once you pick two unique letters, let's say S and T, there are (4!)/(2!*2!) = 6 ways to arrange the four items.
Those 6 ways are:- SSTT
- STST
- STTS
- TSST
- TSTS
- TTSS
Listing these is optional.
But it helps confirm things.
The combinatorics calculator link I mention below is very handy to help list out all possible combos. Specifically refer to the "list them" section.
I divided by 2!*2! to correct for erroneous overcounting.
There are 6 ways to permute SSTT and 3 ways to pick the two unique letters.
We have 6*3 = 18 ways to do case (iii).
Let z = 18 so we can use it later.
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Case (iv)
The five unique letters are: S, T, A, I, C
Use the nPr formula to find there are 5P4 = 120 different permutations where we select 4 items from a pool of 5 (order matters).
I'll leave the scratch work steps for the student to do.
Let w = 120 so we can use it later.
In case (iv) we do not have to worry about repeat elements since we are selecting one item each from {S, T, A, I, C}
-------------------------------------------------------------
We found that
x = 32
y = 216
z = 18
w = 120
which represent the number of ways to do case (i) through case (iv).
Since each case is disjoint, i.e. no overlap, we add up those sub-results to get the grand total.
x+y+z+w = 32+216+18+120 = 386 is the final answer
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How to verify this? You could do so by hand.
That might take a while and there's a possibility of error.
Software is the much better choice.
Luckily there's a combinatorics calculator that specializes in this sort of thing.
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html
The inputs are marked in blue
Names: short
Number of types = n = 10
Number chosen = r = 4
Order important? Yes
Repetition allowed? No
In the "List Them" section, type S,T,A,T,I,S,T,I,C,S
The commas are very important. Do not delete duplicate input letters.
Then click the "list" button underneath that box. It mentions there are 5040 entries. But the implication is that the 5040 is only if we could tell the repeats apart.
After removing the duplicate outputs, there are 386 unique outputs. Each item is listed out so you can check it yourself.
More practice with a similar question
https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1192625.html
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
There are 5 different letters: 3 each of S and T, 2 of I, and 1 each of A and C.
To choose 4 letters, we have the following different cases to consider:
(1) 3 of one letter and 1 of a different letter
(2) 2 each of two different letters
(3) 2 of one letter and 1 each of two other letters
(4) 4 different letters
(1) 3 and 1. We need to...
choose one of the two letters that occurs 3 times: C(2,1)=2
choose one of the other 4 letters: C(4,1)=4
arrange the 4 letters in any of  ways
Number of arrangements: (2)(4)(4) = 32
(2) 2 and 2. We need to...
choose two of the three letters that occurs 2 or more times: C(3,2)=3
arrange the 4 letters in any of  ways
Number of arrangements: (3)(6) = 18
(3) 2, 1, and 1. We need to...
choose one of the three letters that occurs 2 or more times: C(3,1)=3
choose two of the other 4 letters: C(4,2)=6
arrange the 4 letters in any of  ways
Number of arrangements: (3)(6)(12) = 216
(4) 1, 1, 1, and 1. We need to...
choose any 4 of the 5 letters: C(5,4)=1
arrange the 4 letters in any of 4! = 24 ways
Number of arrangements: (5)(24) = 120
Total number of arrangements: 32+18+216+120 = 386
ANSWER: 386
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