1. ## !!Need someone to check answers - Combinatorics Problems

36. Solve for n.

a) P (n,5) = 20P (n,3)

37. A library shelf contains 7 different books. In how many ways can they be arranged if

a) there are no restrictions

Answer: They can be arranged in 5040 different ways 7! = 5040

b) the Mathematics of Data Management textbook must be first

Answer: 6! x 1 = 720 different ways

c) the 3 mathematics textbooks must be together

Answer: - when the 3 textbooks go together, they can be counted as one textbook
Therefore, 4! x 1 = 24 different ways.

2. Originally Posted by cnmath16
36. Solve for n.

a) P (n,5) = 20P (n,3)

What does P(n,5) mean?

Originally Posted by cnmath16
37. A library shelf contains 7 different books. In how many ways can they be arranged if

a) there are no restrictions

Answer: They can be arranged in 5040 different ways 7! = 5040

b) the Mathematics of Data Management textbook must be first

Answer: 6! x 1 = 720 different ways
a) is correct

b) is correct

Originally Posted by cnmath16
c) the 3 mathematics textbooks must be together

Answer: - when the 3 textbooks go together, they can be counted as one textbook
Therefore, 4! x 1 = 24 different ways.

Well, and 4! means, that you have this event:

Mathbooks, Book A, Book B, Book C, Book D

Mathbooks, Book A, Book C, Book B, Book D

...

Book A, Book A, Book C, Book B, Book D?

There are

Mathbooks, A,B,C,D
A,Mathbooks,B,C,D
A, B, mathbooks, C, D
A,B, C, mathbooks, D
A,B,C,D, mathbooks

5 different ways, thats why I think the soluition is 4!*5

3. 37. c) Answer: 5! x 3!

Do you see why?

4. Originally Posted by awkward
37. c) Answer: 5! x 3!

Do you see why?
Yes i do.
First: i wrote

Mathbooks, A,B,C,D
A,Mathbooks,B,C,D
A, B, mathbooks, C, D
A,B, C, mathbooks, D
A,B,C,D, mathbooks

i forgot, that there are no restriction, means for A,B,C,D there are 4 ! possobilities.

Adding the mathbooks you got A,B,C,D,math => 5!

there are three mathbooks
possible orders are

math1, math2, math3
math1, math3, math2
math2, math3, math 1
math2, math1, math 3
math3, math1, math 2
math3, math2, math 1

=> 3!

=> 5! times 3!