Question

A total of $5$ different mathematics books, $4$ different physics books and $2$ different chemistry books are to be arranged in a row in a book shelf. Which of the following is (are) TRUE?

• A. The number of arrangements in which two chemistry books are separated is $9\times 10!$
• B. The number of arrangements in which four physics books are together is $8!4!$
• C. The number of arrangements in which no two mathematics books are together is $(7\cdot 6)(6!)$
• D. The number of arrangements in which the books of the same subject are all together is $12(4!\cdot 5!)$

Answer 1

Answer: (A), (B), (D)

The correct options are
A The number of arrangements in which two chemistry books are separated is $9\times 10!$
B The number of arrangements in which four physics books are together is $8!4!$
D The number of arrangements in which the books of the same subject are all together is $12(4!\cdot 5!)$

Total number of arrangements $=11!$
Number of arrangements that two chemistry books are together $=2!\times 10!$
Hence, number of arrangements that two chemistry books are separated $=11!-2!\times 10!=9\times 10!$

Number of arrangements in which four physics books are together $=8!4!$ since four physics books can also be permuted among themselves in $4!$

Similarly, number of arrangements in which the books of the same subject are all together $=3!(5!4!2!)=12(4!\cdot 5!)$

We want no two mathematics books should be together.
So, let us first place the other six books.
$\_\_\_\_\_\_\_\_\_\_\_\_$
This can be done in $6!$ ways as there are six blanks.
Now, we can put mathematics books between these $7$ gaps.
This can be done in ${}^7{C}_5\times 5!$
Hence, number of arrangements in which no two mathematics books are together is $6!\times {}^7{C}_5\times 5!$