Question 1151101: How many words can be formed from the word SAMARINDA, if
i)4 letters are to be chosen
ii) only 3 letters can be used
Thank you!
Found 3 solutions by Edwin McCravy, jim_thompson5910, ikleyn:
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website! How many words can be formed from the word SAMARINDA, if
i)4 letters are to be chosen
The letters in alphabetical order are AAADIMNRS
Case 1: All 4 letters come from {D,I,M,N,R,S}
6P4=360 ways.
Case 2: Exactly 1 A and 3 letters from {D,I,M,N,R,S}
An example would be SAND.
Choose the 3 non-A letters from {D,I,M,N,R,S} in 6C3=20 ways.
Arrange the 4 letters in 4P4=4!=24 ways.
That's 20∙24=480 ways for case 2.
Case 3: Exactly 2 A's and 2 letters from {D,I,M,N,R,S}.
An example would be ADAM
Choose the 2 non-A letters from {D,I,M,N,R,S} in 6C2=15 ways
Arrange the 4 letters in 4!/2!=24/2=12 ways
That's 15∙12=180 ways for case 3.
Case 4: Exactly 3 A's and 1 letter from {D,I,M,N,R,S}.
An example would be ABAA
Choose the 1 non-A letters from {D,I,M,N,R,S} in 6C1=6 ways
Put it in any of 4 positions.
That's 6∙4=24 ways for case 4.
Total for the three cases: 360+480+12+24=1044
ii) only 3 letters can be used.
Case 1: Exactly 0 A's and all 3 letters from {D,I,M,N,R,S}
An example would be RIM.
Choose and position the 3 non-A letters from {D,I,M,N,R,S} in 6P3=6∙5∙4=120
ways.
That's 120 ways for case 1.
Case 2: Exactly 1 A and 2 letters from {D,I,M,N,R,S}.
An example would be AIR
Choose the 2 non-A letters from {D,I,M,N,R,S} in 6C2=15 ways
Arrange the 3 letters in 3!=6 ways
That's 15∙6=90 ways for case 1.
Case 3: Exactly 2 A's and 1 letter from {D,I,M,N,R,S}.
An example would be ADA
Choose the 1 non-A letter from {D,I,M,N,R,S} in 6C1=6 ways
Put it in any of 3 positions.
That's 6∙3=18 ways for case 1.
Case 4: Exactly 3 A's
The only one is AAA.
That's 1 way for case 4
Total for the four cases: 120+90+18+1=229
Edwin
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website!
Part (i)
We need to break things down into four cases:
- No A's are selected
- Exactly 1 A is selected
- Exactly 2 A's are selected
- Exactly 3 A's are selected
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Case A) We pick no A's.
We have 9 letters, 3 of which are A. There are 6 letters which arent A.
We have 6 letters to pick from for the first slot, then 5 for the next, and so on until four slots are filled. Multiply out the values:
There are 6*5*4*3 = 360 different permutations in which we pick four items from the set {S, M, R, I, N, D}
This is the same as 6 P 4, which is permutation notation.
We will use the value 360 later, so let A = 360.
---------------
Case B) We pick exactly 1 A.
There are 4-1 = 3 slots left to fill. To fill those slots, we pick from this set {S, M, R, I, N, D}
So let's say we have
SAMR
as one of the four letters we pick. There are 4! = 24 ways to arrange those 4 letters.
There are 6 C 3 = 20 different groups of three letters picked from the set {S, M, R, I, N, D}.
So we really have 24*20 = 480 different ways to pick exactly 1 A and 3 other letters, then rearranging those four selections.
We will use the value 480 later, so let B = 480.
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Case C) We pick exactly 2 A's.
Let's dedicate the first two slots to those A's
The other two slots are filled with two letters picked from {S, M, R, I, N, D}
There are 6 C 2 = 15 groups possible.
After we pick those 2 letters, we then have a group of 4. So let's say we pick M and R from the list. That means we have
AAMR
There are 4! = 24 ways to arrange this group, but the 2 A's are indistinguishable. So we must divide by 2.
Overall, there are 15*24/2 = 180 different ways to pick two A's, two other letters, and rearrange the four items.
We will use the value 180 later, so let C = 180.
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Case D) We pick exactly 3 A's (ie all of the A's have been picked)
The first three slots are dedicated to A
We have 6 choices to fill that fourth slot
So we have something like AAAM
For any one letter that isnt "A", there are 4 slots that it could go. Since there are 6 letters that aren't A, this means there are 4*6 = 24 different permutations in which we have selected 3 A's and 1 other letter.
We will use the value 24 later, so let D = 24.
--------------
Add up the results from Cases (A) through (D)
A+B+C+D = 360+480+180+24 = 1044
Answer: 1044
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Side note:
You can use this calculator
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html
To generate all 1044 permutations in which duplicate entries are tossed out.
These are the inputs I gave to the calculator
- names = short
- n = types to choose from = 9
- r = number chosen = 4
- Order Important? Yes
- Repetition Allowed? No
- In the "List them" box, I typed in "S, A, M, A, R, I, N, D, A" without quotes
Ignore the 3024 as that would only apply if we didn't have any repeated letters.
This is what the output should look like after you click the "list" button.
I added the red arrow to help draw your attention to the final answer.
===================================================
Part (ii)
The same idea is used back in part (i) above, but we only consider 3 letters
---------------------
Case A) No A's are selected
6 P 3 = 120 different permutations here
since there are 6 items that are not A, and there are 3 slots to fill.
Let A = 120 so we can use this later.
---------------------
Case B) Exactly 1 A is selected.
Lock down slot 1 for the letter A.
The other 2 slots have 6 C 2 = 15 choices.
There are 3! = 6 ways to arrange any one group of three letters.
There are 15*6 = 90 ways to have a permutation consisting of exactly 1 A, 2 other letters.
Let B = 90 so we can use it later.
---------------------
Case C) Exactly 2 letter A's are selected.
Reserve the first two slots for A's. There is one slot left to fill. There are 6 letters to choose from. One such permutation is AAM. There are 3 ways to arrange the letters in this particular string.
So there are 3*6 = 18 ways to have a permutation consisting of 2 A's and 1 other letter.
Let C = 18 so we can use it later.
---------------------
Case D) Exactly 3 letter A's are selected.
There is only one way to do this and that is AAA. Not much to this part.
Let D = 1.
---------------------
Add up the results of each case
A+B+C+D = 120+90+18+1 = 229
Answer: 229
To check your work, you can use the same calculator as mentioned in part (i). The only change you need to do is make r = 3 (instead of r = 4). The other inputs will remain the same.
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
The solution and the analysis method by Edwin are very nice (!)
I like it very much.
Unfortunately, at the very end of the solution to the part ii), Edwin made an arithmetic mistake (typo ? )
by getting 120+90+18+1=221, while the correct sum is 229.
Fortunately, Jim noticed this mistake and fixed it.
Learn the method from both authors (!)
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