Question 1094038
<br>I will assume you mean arrange them on a shelf, in a single row....<br>
{{{(5+4+6+1+1+1+1+1+1+1+1)!/((5!)(4!)(6!)(1!)(1!)(1!)(1!)(1!)(1!)(1!)(1!)) = 23!/((5!)(4!)(6!))}}}<br>
It's a general principle; let's look at a simple example: 3 copies of one book ("A"), 2 of another ("B"), and 1 of a third ("C").<br>
If all 6 books were different, the number of ways of arranging them would be 6!:
{{{6!}}}<br>
Imagine a list of all those 6! ways of arranging the 6 books.  The 3 A books can be ordered in 3! different ways.  That means every entry in the list occurs 3! times; so the list is too large by a factor of 3!.  So the number of arrangements is now
{{{6!/3!}}}<br>
Likewise, the 2 B books can be arranged in 2! different ways, so again the list is too large by a factor of 2!; the number of distinct arrangements is now 
{{{6!/((3!)(2!))}}}<br>
And since there are no more books with multiple copies, that is the final number of distinct arrangements.<br>
In your problem, there are a total of 23 books, with 6 of one, 5 of another, and 4 of a third, with the rest being single books.  Thus the number shown at the beginning of my response.