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Janet has $10$ different books that she is going to put on her bookshelf. Of these, $4$ are Book C, $3$ are Book B, $2$ are Book S and $1$ Book P. Janet wants to arrange her books so that all the books dealing with the same subject are together on the shelf. How many different arrangements are possible?
My workings are
$4! \cdot 3! \cdot 2! \cdot 1!$
but the actual answer has an additional multiplication by $4!$
Why is this so? My guess is because we need the $4$ books to come together, let's say for book C. But isn't that "shown" by the working of $4!$ for Book C and $3!$ for Book B and so on?
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Madi K.
asked • 04/10/191 Expert Answer
This is an example of a permutation, because the order of the books matters. If the order didn't matter, it would be a combination.
The formula for the number of permutations of n objects taken r at a time is given by this formula:
P(n,r)=n!/(n-r)!
where ! is the factorial operator. In our example, n=5 and r=3, so we have:
P(5,3) = 5!/(5-3)!
= 5!/2!
= 5*4*3*2*1/2*1
= 5*4*3
= 60