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.