CS计算机代考程序代写 discrete mathematics The University of Melbourne — School of Mathematics and Statistics

The University of Melbourne — School of Mathematics and Statistics

MAST30012 Discrete Mathematics — Semester 2, 2021

Practice Class 2: Arrangements and Combinations – Answers

Q1: (a)


20

3


=

20!

3! 17!
= 1140.

(b) 7 · 5 · 3 · 1 = 105.

(c)


8

2

� �
6

2

� �
4

2

� �
2

2

4!
=

8!

24 4!
=

8!

8 · 6 · 4 · 2
= 7 · 5 · 3 = 105.

(d) Total number of possible configurations is 210. 5 heads can be chosen in

10

5


ways so

Pr(5 heads) =


10

5

210
=

63

256
= 0.246073 . . .

At least 5 heads in a row. Number of favourable outcomes is 25+5⇥ 24 = 112 and hence

Pr(at least 5 heads in a row) =
112

1024
=

7

64
= 0.109375.

Exactly 5 heads in a row. N umber of favourable outcomes is 64 and therefore

Pr(exactly 5 heads in a row) =
1

16
= 0.0625.

Q2: (a) aaa, aab, abb, bbb total of 4 =

2+3�1

3


=


4

3


.

(b)


20 + 3� 1

3


=


22

3


=

22!

3! 19!
= 1540.

(c) Number of ways to arrange n� 1 symbols I, and r symbols in a line is

n+r�1

r

(d) Same as (c).

Q3: (a) Derivation required.

(b) Derivation required.

(c) Arguments required.

Q4: (a) Derivation required.

(b) Derivation required.

Q5: (a) There are

n
r


ways to order r ‘1’s and n� r ‘2’s in a line.

(b) By the stated correspondence (bijection) the two counting problems are the same.