Question 1151101
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Part (i)


We need to break things down into four cases:
<ol type="A"><li>No A's are selected</li><li>Exactly 1 A is selected</li><li>Exactly 2 A's are selected</li><li>Exactly 3 A's are selected</li></ol>


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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.



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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.



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Add up the results from Cases (A) through (D)
A+B+C+D = 360+480+180+24 = 1044


<font color=red size=4>Answer: 1044</font>



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Side note:


You can use this calculator
<a href = "https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html">https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html</a>
To generate all 1044 permutations in which duplicate entries are tossed out. 



These are the inputs I gave to the calculator
<ul><li>names = <font color=blue>short</font></li><li>n = types to choose from = <font color=blue>9</font></li><li>r = number chosen = <font color=blue>4</font></li><li>Order Important? <font color=blue>Yes</font></li><li>Repetition Allowed? <font color=blue>No</font></li><li>In the "List them" box, I typed in "<font color=blue>S, A, M, A, R, I, N, D, A</font>" without quotes</li></ul>
<img width = "50%" src = "https://i.imgur.com/Q8CKgM9.png">
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.
<img width = "50%" src = "https://i.imgur.com/R5dQuXc.png">
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


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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.


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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.


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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.

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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.
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Add up the results of each case
A+B+C+D = 120+90+18+1 = 229


<font color=red size=4>Answer: 229</font>


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. 
<img width = "50%" src = "https://i.imgur.com/UQs6VhP.png">
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