A student has 4 English books,3 math books,and 5 history books.In how many ways can these books be arranged on a shelf if the books in same subjects must be kept together?
There are 4!, 3!, and 5! ways to arrange each subject
Since there are 3! ways to arrange the order of the subjects, the total ways is
4!3!5!3! = 103680
You all are terrible at this and Steve is so wrong like all the time.
3*4*5 = 60
To solve this problem, we need to consider three separate groups - English books, math books, and history books. Since books of the same subject need to be kept together, let's first calculate the number of ways we can arrange each group individually.
For the English books, there are 4 books available, so they can be arranged in 4! (4 factorial) ways, which equals 4 x 3 x 2 x 1 = 24 ways.
Similarly, for the math books, there are 3 books available. Therefore, they can be arranged in 3! (3 factorial) ways, which equals 3 x 2 x 1 = 6 ways.
And for the history books, there are 5 books available. So, they can be arranged in 5! (5 factorial) ways, which equals 5 x 4 x 3 x 2 x 1 = 120 ways.
Now, we need to consider the three groups as a single entity and find the number of ways to arrange them. Since there are three groups, we can arrange them in 3! (3 factorial) ways, which equals 3 x 2 x 1 = 6 ways.
To find the final result, we multiply the number of ways to arrange each group together: 24 x 6 x 120 x 6 = 207,360 ways.
Therefore, there are 207,360 ways to arrange the books on the shelf if the books of the same subject must be kept together.