How many ways can you arrange four different Mathematics books and two different Science books on the shelf such that the Scimo books must be the first ar last book?
Answers
Step-by-step explanation:
We are given four different mathematics books, six different physics books and two different chemistry books.
A) There are two parts to this question. In the first part, we will look at the subject books as a whole. Since we are asked to keep the books in a particular subject together, we will assume all the mathematics books to be one single element, all the physics books as one element and all the chemistry books as one element. Now, we have three different elements which can be arranged in 3!3! ways.
Now, within each single element of these books, there are the actual number of books. For example, in one element in mathematics, there are four books. These books will arrange within themselves in 4!4! ways.
Similarly, in one element of physics, there are six books. The books will arrange within themselves in 6!6! ways.
In one element of chemistry, there are two books which will arrange within themselves in 2!2! ways.
Therefore, total possible arrangements= 3!×4!×6!×2!3!×4!×6!×2! = 6×24×720×2=2,07,3606×24×720×2=2,07,360ways.
B) In this part, we are only required to keep all the math books together. We will follow the same approach as above and consider all the mathematics books as one single element. Now, there are 6 physics books, 2 chemistry books and 1 element of math books. In total, we have 6+2+1=96+2+1=9 books now.
These 9 books can be arranged in 9!9! ways. After this, we will look at that one element of math books. These books will arrange themselves in 4!4! ways because there are 4 books.
Now, total arrangements = 9!×4!=3,62,880×24=87,09,120
Answer:
Assumption: the books don’t have to be grouped together (You could have MMSMESMSE for example), but each amongst each “type” of book they are nondistinct (ie - every math book is identical, etc).
This would be done the same way as a “how many ways are there to arrange the word mmmmsssee” type problem: 9!4!3!2!=1,260 arrangements