5different mathematics books, 4 different french books and 2 different biology books are to be arrange on a shelf. how many different arrangements are possible, if
1. the books in each particular subject must all stand together?
2. only the mathematics books must stand together.

Problem 1

A = 5! = 5*4*3*2*1 = 120 ways to arrange the math books

B = 4! = 4*3*2*1 = 24 ways to arrange the French books

C = 2! = 2*1 = 2 ways to arrange the biology books

D = 3! = 3*2*1 = 6 ways to arrange the three blocks of books, each block a different subject

E = A*B*C*D = 120*24*2*6 = 34560 different ways to arrange the books such that the subjects stick together

Problem 2

Group all the math books to form one "book". This is the same as taking out the 5 math books and replacing it with one single math book. This single math book acts as a placeholder for all of the five books together as one block.

We have

1 math "book"

4 French books

2 biology books

So there are 1+4+2 = 7 "books" total and

7! = 7*6*5*4*3*2*1 = 5040 different ways to arrange those "books"

Within the math block of books, there are 5! = 120 ways to arrange things.

So overall we have 120*5040 = 604800

well we know there are 4 clubs in 52 cards so it’s replaced so it should be 3/51

step-by-step explanation:

isosceles obtuse

step-by-step explanation: