# How many ways can $2$ different history books, $5$ different math books, and $4$ different novels be arranged on a shelf if ...?

This is my first class in probability so I just wanted verification as to my attempted solution.

Question:

In how many ways can $$2$$ different history books, $$5$$ different math books, and $$4$$ different novels be arranged on a shelf if the books of each type must be together?

Solution:

$$2! \cdot 5! \cdot 4! \cdot 3! = 34560$$

There are $$34560$$ ways the books can be arranged.

• Looks fine to me... These days people put too much stress on just the numeric result being right, without the "show your works" bit: I have a guess that you probably understand this matter and can explain where those $2!$, $5!$, $4!$ and $3!$ are coming from, but it would be better if you explicitly spelled them in your solution. ("First, we have $3!$ ways to decide the order of the groups of the books of the same type. Then, for every one of those ways, we can order the history books in their group in $2!$ ways etc. etc.") Sep 6 '19 at 20:31
• Thank you for the constructive feedback. I will definitely do that in the future.
– Mark
Sep 6 '19 at 20:32

Question: "In how many ways can 2 different history books, 5 different math books, and 4 different novels be arranged on a shelf if the books of each type must be together?"

In this question, sequence of the books is not important, therefore:

• For the 2 history books: 2 ways to arrange them (AB and BA), or $$2!$$
• For the 5 math books: $$5*4*3*2*1 = 5!$$ ways to arrange them, or 120
• For the 4 novels: $$4*3*2*1 =$$4!\$ ways to arrange them, or 24

Think like this:

• For the history books (assuming we only look at the history books): 2 options for the first slot, and 1 for the last
• For the math books (again, only look at the math books): 5 options for the first slot, $$5-1=4$$ for the second slot, $$5-2=3$$ for the third and so on
• The same for the novels

We also have three types of books, so, the order of first-to-appear is, by the same logic, 3!

Therefore, in the the end you have $$2!*5!*4!*3!=34560$$ ways to arrange those books