Combinations
By Mr Porter
Definition: Combination Selections.
The number of unordered selections of r items form n items (that is the number of
combinations of r items from n items) is given by
n
Cr =
n!
r!(n - r)!
r - number of item you want to select.
n - number of items you are selecting from.
Alternate Notation:
æ n ö
n!
=
çè r ÷ø r!(n - r)!
In this powerpoint, nCr , notation will be used.
Example 1
How many subgroups of 3 can be selected from a group of 10 people?
The selection of the subgroup of 3 can be in any order  order is not important.
Un-order Selected = 10C 3
10!
3!(10 - 3)!
10!
=
3!7!
=
= 120
Use a scientific Calculator to evaluate.
Students should practice the expansion simplification.
n
Cr =
n!
r!(n - r)!
Example 2
A university student can borrow 4 books from the library. She wants to read 3 Biology
books, 2 Mathematics, 4 Chemistry and 3 Physics books.
a) How many different selections of 4 books can be made?
b) How many selections can she make if she must takes only 2 Chemistry books?
c) How many selections can be made if she takes at least 2 Physics books?
a) Different selections of any 4 books.
c) She takes at least 2 Physics books
The total number of books = 3 + 2 + 4 + 3 =12.
Selections =
=
12C
4
n
12!
4! 8!
She can takes at 2, 3 or 4 Physics Books.
n!
Cr =
r!(n - r)!
2 Physics Books = 4C2
3 Physics Books = 4C3
4 Physics Books = 4C4
Selections = 4C2 x 8C2 + 4C3 x 8C1 + 4C4 x 8C0
= 495
b) Selections takes only 2 Chemistry books
Select 2 of the Chemistry books first = 4C2
Select 2 other books from the rest of the not including
the remaining Chemistry books = 8C2
Selections = 4C2 x 8C2
4!
8!
=
´
2!2! 2!6!
= 168
=
4!
8!
4!
8!
4!
8!
´
+
´
+
´
2!2! 2!6! 3!1! 1!7! 4!0! 0!8!
= 201
Example 3
A team of 3 students is to be chosen from a group of 8 girls and 5 boys to form a debating
team. How many different teams can be chosen:
a) without restrictions.
b) if the the team is to contain a majority of girls.
a) Team without restrictions
Total number of students = 8 + 5 = 13
n!
n
team of 3 students = 13C3 Cr =
r!(n - r)!
13!
=
3!10!
= 286
b) team is to contain a majority of girls
There could be 2 or 3 girls in the team.
2 girls + 1 boy = 8C2 x 5C1
3 girls + 0 boys = 8C3 x 5C0
majority of girls = 8C2 x 5C1 + 8C3 x 5C0
=
8!
5!
8!
5!
´
+
´
2!6! 1!4! 3!5! 0!5!
= 186
Example 4
In how many ways can a committee of 4 women and 3 men be chosen from 8 women
and 7 men?
a) Without restrictions?
b) If Sue refuses to serve if Betty is selected?.
a) Without restrictions
b) Sue refuses to serve if Betty
Select from each group, women then men.
Committee of 4W + 3M = 8C4 x 7C3
8!
7!
=
´
4!4! 3!4!
= 2450
Select Betty and Sue plus 2W = 2C2 x 6C2.
Select 3M remains the same = 7C3
Committee without BOTH Sue and Betty
= unrestricted – (Betty and Sue) selected
= 8C4 x 7C3 – 2C2 x 6C2 x 7C3
æ 8!
7! ö æ 2
6!
7! ö
=ç
´
´
´
è 4!4! 3!4!÷ø çè 2!0! 2!4! 3!4!÷ø
= 1925
Example 5
In person must answer exactly 7 of 10 questions in an examination. Given that they must
answer at least 3 of the first 5 questions, find the number of ways in which 7 questions can
be selected.
The person can select 3 , 4 or all 5 from the first 5 questions and MUST choose 4, 3 and 2 from the remaining 5 questions.
exactly 7 of 10 questions = 5C3 5C4 + 5C4 5C3 + 5C5 5C2
=
5!
5!
5!
5!
5!
5!
´
+
´
+
´
3!2! 4!1! 4!1! 3!2! 5!0! 2!3!
= 110
Example 6
Find the number of different, selections of 5 letters that can be made from the
letters of the word SYLLABUS.
The word SYLLABUS has 8 letters, made up of
2 ‘L’, 2’S’ and 4 non-repeating letters {Y, A, B, U}.
We can select 5 letters as follows:
a) 2L, 2S, 1 from {Y, A, B, U}
= 2 C2 2 C2 4 C1
b) 2L, 3 from {S, Y, A, B, U}
= 2 C2 5 C3
c) 2S, 3 from {Y, L, A, B, U}
= 2 C2 5 C3
d) 5 from {S, Y, L, A, B, U}
= 6 C5
Use a scientific Calculator to evaluate.
Number of 5 letters that ‘words’= 2C2 2C2 4C1 + 2C2 5C3 + 2C2 5C3 + 6C5
2!
2!
4!
2!
5!
2!
5!
6!
=
´
´
+
´
+
´
+
2!0! 2!0! 1!3! 2!0! 3!2! 2!0! 3!2! 5!1!
= 30
Example 7
Find the number of different, selections of 4 letters that can be made from the
letters of the word CONCYCLIC.
The word CONCYCLIC has 9 letters, made up of
4 ‘C’ and 5 non-repeating letters {O, N, Y, L, I}.
We can select 4 letters as follows:
a) 4 C, 0 from {O, N, Y, L, I} = 4C4 5C0
b) 3 C, 1 from {O, N, Y, L, I} = 3C3 5C1
c) 2 C, 2 from {O, N, Y, L, I} = 2C2 5C2
d) 1 C, 3 from {O, N, Y, L, I} = 1C1 5C3
e) 0 C, 4 from {O, N, Y, L, I} = 0C0 5C4
Use a scientific Calculator to evaluate.
Number of 4 letters selections = 4C4 5C0 + 3C3 5C1 + 2C2 5C2 + 1C1 5C3 + 0C0 5C4
4!
5!
3!
5!
2!
5!
1!
5!
0!
5!
=
´
+
´
+
´
+
´
+
´
4!0! 0!5! 3!0! 1!4! 2!0! 2!3! 1!0! 3!1! 0!0! 4!1!
= 31