Question

19 1/3 Homework 1: Problem 3 b) the Answer (c) the attempts remaining.

Answer 1

(a)

Total different books =5+4+1 =10.

Number of ways of arranging 10 different books in any order =10P₁₀ =10! =10*9*8*....*2*1 =3,628,800 ways

(b)

Treating mathematics books and novels as 2 units. We have 2 units+1 biology book =3 units. These 3 units can be arranged in 3P₃ =3! =6ways. 5 novels can be arranged among themselves in 5P₅ =5! =120 ways. 4 mathematics books can be arranged among themselves in 4P₄ =4! =24 ways.

So, total number of ways of arranging books so that 5 novels are together and 4 mathematics books are together =6*120*24 =17,280 ways. ​​​​​​

(c)

Treating 4 mathematics books as unit, we have 1 unit+5 novels+1 biology book =7 units which can be arranged in 7! ways​​​​​​. 4 mathematics books can be arranged among themselves in 4! ways​​​​​​. So, total number of ways of arranging books so that mathematics books are together =7!*4! =5,040*24 =120,960 ways. ​​​​​​