10b)

Consider the following cases :

① one person gets all 5 books.
Number of ways = 10C1 = 10

② 1 person gets 4 books, another gets the last book

Permutations :
1,4
4,1


Number of ways
= 10C2 x 5C4 x 2 permutations
= 450

③ 1 person gets 3 books , one person
gets 2 books

Permutations :
3,2
2,3

Number of ways = 10C2 x 5C3 x 2 permutations = 900

④ 3 persons, one gets 3, the other two 1 each

Permutations :
3 1 1
1 1 3
1 3 1

Number of ways = 10C3 x 5C3 x 2C1 x e permutations = 7200

⑤ 3 persons, one gets 1, the other two get 2 each.

Permutations :
2,2,1
1,2,2
2,1,2

Number of ways = 10C3 x 5C1 x 4C2 x 3 permutations = 10800

⑥ 4 persons, one gets 2 , the other three get one each.

Permutations :
2,1,1,1
1,2,1,1
1,1,2,1
1,1,1,2


Number of ways
= 10C4 x 5C2 x 3C1 x 2C1 X 4 permutations = 50400

⑦ 5 persons , 1 book each.

1,1,1,1,1

Number of ways = 10C5 x 5!(since all books are different) = 30240

Total = 30240 + 50400 + 10800 + 7200 + 900 + 450 + 10 = 100000




Alternatively, a much faster method is.:

Number of ways to give the
1st book = 10
2nd book = 10
3rd book = 10
4th book = 10
5th book = 10

Total number of ways
= 10 x 10 x 10 x 10 x 10
= 10^5
= 100000

12)

105840 = 2⁴ x 3³ x 5¹ x 7²


For the prime factors , we have the following to choose from :

2°,2¹,2²,2³,2⁴ (number of ways = 5)
3°,3¹,3²,3³ (number of ways = 4)
5°,5¹ (number of ways = 2)
7°,7¹,7² (number of ways = 3)

Total number of ways = 5 x 4 x 2 x 3
= 120

However, this includes the factor 1 and 105840. So we need to exclude them.

Number of ways = 120 - 2 = 118