Question

The number of ways that 5 mathematics, 3 physics, and 2 chemistry books can be arranged so that the three physics books kept together and the two chemistry books not together are
A. 181440
B. 180430
C. 118316
D. None

Answer 1

Hint: We will first consider that the three physics books are kept together so, we will tie up three books and we can do it in \[3!\] ways. Now, we are left with 5 math books, 2 chemistry, and 1 bundle of physics books. Here, we will first arrange the math and physics books that are in \[6!\] ways. After arranging these 6 objects, we have got 7 gaps so, we need to assure that chemistry books are placed such that they are not together. So, we can do it in \[{}^7{C_2}\] ways. Thus, we will multiply all the obtained values and simplify it.

Complete step by step answer:

The aim is to find ways such that 5 mathematics, 3 physics, and 2 chemistry books can be arranged in a way that physics books are always together and chemistry books are apart from each other.
We will start by tying up the three physics books such that they are always together and consider them as a bundle.
As the three physics books can do their arrangements in \[3!\] ways so, we conclude that the physics books have \[3!\] ways.
Now, we will arrange the math and physics books so, there are 5 math books and 1 bundle of physics books so we arrange these two sets in \[6!\] ways.
Now, after arranging these 6 objects we have got the 7 gaps as the chemistry books do not have to be together.
So, by arranging the chemistry books in these gaps we have to make sure that no two chemistry books are placed together, so, we can arrange them in \[{}^7{C_2}\] ways.
Thus, by using the fundamental principle of multiplication, we get that the total number of possible ways is \[ \Rightarrow 3! * 6! * {}^7{C_2}\]
Now, we will simplify the above obtained expression using the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] and we get,
\[ \Rightarrow \left( {3!} \right) * \left( {6!} \right) * \dfrac{{7!}}{{2! \times 5!}} = \dfrac{{\left( {3 * 2!} \right) * \left( {6 * 5!} \right) * 7!}}{{2! * 5!}} = 7! * 3 * 6\]
We can further simplify the expression by opening the factorial as \[n! = n \times \left( {n - 1} \right) \times \left( {n - 1} \right) \times ..... \times 2 \times 1\] and we get,
\[ \Rightarrow 90720\]
Thus, we can conclude that there are 90720 possible ways to arrange the books.
Hence, option D is correct.


Note: As there are 3 books of physics so, we can arrange it in \[3!\] ways in the bundle itself. As there are 7 slots in total and 2 books need to be arranged so we have used the combination concept and solve \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]. As the physics books have to be together so it is necessary to tie them up and make a bundle of three books. Do the calculations and substitution of the values in the formula properly. Remember how to open the factorial of \[n\] as \[n! = n \times \left( {n - 1} \right) \times \left( {n - 1} \right) \times ..... \times 2 \times 1\].