
Eleven books consisting of 5 mathematics, 4 physics and 2 on chemistry are placed on a shelf at random. Find the number of possible ways of arranging them on the assumption that the books on the same subject are all together.
Answer
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Hint:
We can consider the books of each subject as one object and find the possible way of arranging them. Further we can find the number of ways in which the book on each of the subjects can be arranged. Then we can take the product of all these arrangements to get the required number of arrangements
Complete step by step solution:
It is given that there are 11 books. Out of which 5 are mathematics, 4 are physics and 2 are chemistry.
We need to arrange these books in such a way that all the books on the same subject are together. For that we can consider the books of each subject as one object.
As there are 3 subjects, they can be arranged in $3!$ ways.
Now books on each subject can be arranged within themselves.
As there are 5 books on mathematics, they can be arranged in $5!$ ways.
As there are 4 books on physics, they can be arranged in $4!$ ways.
As there are 2 books on chemistry, they can be arranged in $2!$ ways.
So, the total possible arrangement is given by the product of all the above arrangements.
$ \Rightarrow A = 3! \times 5! \times 4! \times 2!$
On expanding the factorials, we get,
$ \Rightarrow A = 3\, \times 2 \times 5 \times 4 \times 3\, \times 2 \times 4 \times 3\, \times 2 \times 2$
On clubbing the factors, we get,
$ \Rightarrow A = 5 \times {4^2} \times {3^3} \times {2^4}$
On simplification we get,
$ \Rightarrow A = 5 \times 16 \times 27 \times 16$
On multiplication we get,
$ \Rightarrow A = 34560$
So, the number of ways the book can be arranged such that the books on the same subject are all together is 34560.
Note:
We used the concept of permutations to solve this problem. We cannot use combinations as the order of arranging is important. After considering the books of each subject as one object, we must consider arranging the books in each subject as they are together. This has to be done because the books are not identical.
We can consider the books of each subject as one object and find the possible way of arranging them. Further we can find the number of ways in which the book on each of the subjects can be arranged. Then we can take the product of all these arrangements to get the required number of arrangements
Complete step by step solution:
It is given that there are 11 books. Out of which 5 are mathematics, 4 are physics and 2 are chemistry.
We need to arrange these books in such a way that all the books on the same subject are together. For that we can consider the books of each subject as one object.
As there are 3 subjects, they can be arranged in $3!$ ways.
Now books on each subject can be arranged within themselves.
As there are 5 books on mathematics, they can be arranged in $5!$ ways.
As there are 4 books on physics, they can be arranged in $4!$ ways.
As there are 2 books on chemistry, they can be arranged in $2!$ ways.
So, the total possible arrangement is given by the product of all the above arrangements.
$ \Rightarrow A = 3! \times 5! \times 4! \times 2!$
On expanding the factorials, we get,
$ \Rightarrow A = 3\, \times 2 \times 5 \times 4 \times 3\, \times 2 \times 4 \times 3\, \times 2 \times 2$
On clubbing the factors, we get,
$ \Rightarrow A = 5 \times {4^2} \times {3^3} \times {2^4}$
On simplification we get,
$ \Rightarrow A = 5 \times 16 \times 27 \times 16$
On multiplication we get,
$ \Rightarrow A = 34560$
So, the number of ways the book can be arranged such that the books on the same subject are all together is 34560.
Note:
We used the concept of permutations to solve this problem. We cannot use combinations as the order of arranging is important. After considering the books of each subject as one object, we must consider arranging the books in each subject as they are together. This has to be done because the books are not identical.
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