Question 1191176
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Part (a)


We have 9 books which I'll call 
book a, book b, book c, ..., book h, book i
Let's say we want books a through c to always stick together in any particular order.


We can pull books a,b,c out of the group on a temporary basis for now. Replace them with book j.
The position of book j will represent books a,b,c in any order.
The 9 books drop to 9-3 = 6 after taking books a,b,c out, but then the count bumps up to 6+1 = 7 after adding book j.


Arrange the 7 books and you should find there are 7! = 7*6*5*4*3*2*1 = 5040 permutations. The order matters. 
You can alternatively use the nPr formula with n = 7 and r = 7.


That 5040 describes sequences involving book j. Wherever you see book j, replace it with some permutation of a,b,c.
For example, if we had the sequence
<font color=blue>j</font>,d,e,f,g,h,i
then it could represent any of the following 6 items<ul><li><font color=blue>a,b,c</font>,d,e,f,g,h,i</li><li><font color=blue>a,c,b</font>,d,e,f,g,h,i</li><li><font color=blue>b,a,c</font>,d,e,f,g,h,i</li><li><font color=blue>b,c,a</font>,d,e,f,g,h,i</li><li><font color=blue>c,a,b</font>,d,e,f,g,h,i</li><li><font color=blue>c,b,a</font>,d,e,f,g,h,i</li></ul>The 6 is due to the fact that 3! = 3*2*1 = 6
There are 6 ways to arrange any group of 3 items where order matters.


As another example, the sequence here
g,h,i,<font color=blue>j</font>,d,e,f
could represent any of the following<ul><li>g,h,i,<font color=blue>a,b,c</font>,d,e,f</li><li>g,h,i,<font color=blue>a,c,b</font>,d,e,f</li><li>g,h,i,<font color=blue>b,a,c</font>,d,e,f</li><li>g,h,i,<font color=blue>b,c,a</font>,d,e,f</li><li>g,h,i,<font color=blue>c,a,b</font>,d,e,f</li><li>g,h,i,<font color=blue>c,b,a</font>,d,e,f</li></ul>Overall, there are 6*5040 = <font color=red>30,240</font> ways to arrange the 9 books such that 3 of them are always together in some fashion.


Answer: <font color=red>30,240</font>


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Part (b)


There are 9! = 9*8*7*6*5*4*3*2*1 = 362,880 different ways to arrange all of the books regardless if 3 particular books stick together or not.


We found earlier there are 30,240 ways to arrange the books so that 3 stick together.


Subtract those two values to get
362,880 - 30,240 = <font color=red>332,640</font>
which represents the number of ways to have the 3 books separated in some fashion. Either they are all isolated from one another, or we only have 2 of them together (but not all 3 together).


This works because those 3 books are either always together, or split apart in some way. The two events are complementary of one another.


Answer: <font color=red>332,640</font>

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