BY BENIEL SEKA
Introduction
You
have probably met the term permutation
before. If you have been reading these series of articles, you definitely have
studied it and you are ready to extend its use. The new comers are informed
that a permutation is the number of arrangements possible for an ordered
selection of r objects form n objects. In this lesson you are going to learn a
new term called combination where the
order is not important.
Text
When
the order of selection is not important, that is, when XY and YX mean the same
thing, the number of arrangements is called a combination.
Example 1
A
committee of 2 is chosen from a group of 3 girls. In how many ways can this be
done?
Solution
If
the people are Sia, Rabia and Kaleta, the possible ordered arrangements are:
(Sia
and Rabia), (Sia and Kaleta), (Rabia and Kaleta), (Rabia and Sia), (Kaleta and
Sia), (Kaleta and Rabia). However, a committee of Sia and Rabia is the same as
a committee of Rabia and Sia. The number of unordered arrangements is half the
number of ordered arrangements, in this example, since each committee has 2
ways of being arranged. In fact each has 2! (two factorial) ways of being
arranged.
The
number of ways is 3.
You
can say 
= 
= 3.
= = 3.
Notation
The
number of ways of making unordered selections of r objects from n objects is
denoted by nCr. Recall that the notation nPr
means the ordered selection of r objects from n objects. However, there
are r! ways of arranging the r objects. If order is important, then
nPr
the
unordered selection of r objects is given by r! .
This
is equal to nCr
.
Hence
nCr
= nPr
r!
= 
nCr = 
nCr
can also be written as 
where the notation is known as
‘choose notation’
where the notation is known as
‘choose notation’
Example 2
(a) Evaluate 5C3
S0lution:
nCr = n!
(n-1)!r!
5C3 = 5!
(5-3)!3!
= 5x4x3x2x1
2!3!
=5x4x3x2x1
2x1x3x2x1
= 5x2
= 10
(b) Calculate 4C2
Solution: 4C3
= 4!
(4-3)!3!
=4x3x2x1
1!x3!
= 4x3!
1x3!
=4.
Example 3
a) A
committee of 5 people is formed from 12 people. In how many different ways can
the committee be formed?
b) If
the group consists of 4 men and 8 women, how many ways can the committee be
formed with 2 men and 3 women?
Solution
a) The
order of the committee is not important. For example, a committee of Juma and
Stela is the same as a committee of Stela and Juma. Therefore the number of
committees is 12C5.
12C5 = 
= 
= 
= 
= 792
b) 2
men can be chosen from 4 men in 4C2 ways.
3
women can be chosen from 8 women in 8C3 ways.
Total
number of arrangements = 4C2
X 8C3
= 
X
X
= 
= 
= 8x7x6
=
336
Example
4
a) Ten
people apply for a job but only 5 jobs are available. How many different combinations
are possible for the 5 jobs?
b) Out
of the 10 people, 3 are from Arusha region and 7 are from other regions. If 3
jobs are given to the other regions, how many combinations are possible?
a) Solution 3:
Arusha people can get 2 jobs in 3C2
ways 7 other people for other regions can get 3 jobs in 7C3
ways
Total
combination =
3C2 X 7C3
=
= 
= 105
b)
10C5
= 
= 
=
= 252
The
combination nCr is related to the binomial expansion. For
example,
(a+b)2= 1a2+2ab+1b2
(a+b)3= 1a3+3a2b+3ab2+1b3
(a+b)4=1a4+4a3b+6a2b2+4ab3+1b4
The
coefficients of these expansions can be written as follows:
(a+b)2
=2C0a+2C1ab+2C2b2
(a+b)3 =3C0a3+3C1a2b+3C2ab2+2C3b3
(a+b)4
=4C0a4+4C1a3b+4C2a2b2+4C3ab3+4C4b4
In
general the binomial expansion may be written as:
(
a+b)n = nC0an+nC1an-1b+nC2an-2b2+…+nCnbn.
Example 5
Find
the third term of the expansion of (a+b)15
using the combination notation.
Solution:
n=15
The
expression for the third term is 15C2a13b2= 15!a13b2
(15-2)!2!
= 15x14x13!a13b2
13!2x1
= 15x7a13b2
= 105a13b2.
Exercise
1.
In a class of 40 students, 5 are
selected to participate in a school debate. If the selection is made at random,
find in how many ways the selection is possible.
2. A
committee of 3 women and 5 men is to be selected at random from seven women and
seven men. Find the number of ways that committee can be chosen.
3. In
a group of 5 volunteers in a job, a team of 4 is put in a special job. If the
selection is made at random, in how many different ways can the selection of
the team made?
4. A
competition team of 4 is chosen from 15 players. How many different
combinations are possible:
a) If
there are no restrictions of who is in the team?
b) If
two particular people are to be included? (Ans to Qu. 4: (a) 1365 (b) 78).
5. Expand;
(a) (x+y)5. Answer: x5+5x4y+10x3y2+19x2y3+5xy3+y5
(b) (x-3)4 Answer:
x4-12x3+54x2-108x+81.
6. What
is the coefficient of x2 in the expansion of (2-x)4?
Answer:
24
7.
A city uses 6 digits for its telephone numbers, with the first 2 digits
the same for all the telephone
numbers. Hw many different telephone numbers are possible?
Answer:
10 000.
8. A
team of four boys and five girls is to be selected randomly from a group of
eight male and seven female athletes for Olympic games. If Mr Chakubanga and
Miss Roda are both hoping too be selected,
find the probability that both will be selected. [answer 5/14 ].
-END-
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