Chapter 11 COMBINATIONS
11.1 Introduction: A
combination is an un-ordered
collection of distinct elements, usually of a prescribed size and taken from a
given set. In other words, each of the different group which can be formed by
taking some or all the number of objects, irrespective of ther arrangements is called
as combinations.
For example: A
salad is a combination of different fruits. It doesnt matter what is he order
of fruits. It could be apples, pineapples and tomato or tomato, apple and
pineapple.
There
are also two types of combinations :
·
Repetition is Allowed: such as coins in your pocket (5,5,5,10,10)
·
No Repetition:
such as lottery numbers (2,14,15,27,30,33)
Combination
of n things taken r at a time is represented as
C(n, r) = nCr
C(n,
r) is defined well when n and r are integers such that
n
≥ r
and
r
≥ 0
11.2 Combination of n different objects:
i) The number of all combinations of n
distinct objects taken r at a time is given by;
C(n, r) = n!
r! (n r) !
where
n! is a factorial.
Example: Evaluate
C(10, 2)
Solution: Here,
n = 10
And r = 2
Using
the general formula
C(n, r)
= n!
r! (n r) !
we
get
C(10, 2)
= 10!
2! (10 2)!
=
10 x 9 x 8!
2! 8!
= 45
ii)
C(n, n) = 1
Proof: we
have,
C(n, r) = n!
r! (n r) !
putting
r = n
C(n, r) = n!
n! (n n) !
= n!
n! 0!
we
know that
0! = 1
therefore,
C(n,
n) = 1
iii) C(n,
r) = C(n, n-r) for 0≤ r ≤ n
Proof: C(n,
n-r) = n!
(n r) ! x [n (n r)] !
= n!
(n r) ! x (r) !
= C(n, r)
iv)
C(n, r) + C(n, r 1) = C(n+1, r)
Proof:
C(n, r) + C(n, r 1) = n! +
n!
(n r) ! x (r) ! (r 1) ! x [n (r 1)] !
= n! + n!
< (n r) ! x (r) ! (r 1) ! x [n r +1] !
= (n!). (n r +1) +
n!. r
(r) ! . (n r+1) ! r ! x (n r +1) !
= (n!) + (n r +1+r)
(r) ! . (n r+1)!
= (n+1). n!
(r)
! . (n r+1) !
=
(n+1) !
r! (n+1 r)
!
=
C(n+1, r)
11.3 Practical problems:
Example1: In
how many ways can a cricket team be chosen out of a batch of 20 players, if
i)
there
is no restriction in the selection;
ii)
a
particular player is always been chosen;
iii)
a
particular player is never chosen;
solution: i) the number of ways In which 11 players
can be chosen out of 20 players;
here, n= 20 and r = 11
therefore,
C(20,11) = C(20, 20
11) = C(20, 9)
=
20!
9! (20 9) !
=
20x19x18x17x16x15x14x13x12
9x8x7x6x5x4x3x2x1
=
167960
ii)
when
a particular player is chosen always;
= C(19, 10) = C(19, 9)
= 19 x 18x17x16x15x14x13x12x11
9x8x7x6x5x4x3x2
= 21318
i)
when
a particular player is never chosen;
=
C(19, 11) = (19, 2)
= 19 x 18
2
= 171
Example2: In an examination , a candidate has to pass
in each of the 5 subjects. In how many
ways can he fail.
Solution:
the student can fail by failing in 1 or 2 or 3 or 4 or 5 subjects out of
5 in each case.
Therefore,
the total number of ways in which he can fail
= C(5,
1) + C(5 2) + C(5, 3) + C(5, 4) + C(5, 5)
=
C(5, 1) + C(5 2) + C(5, 2) + C(5
1)
[using C(n, r) = C( n, n- r)]
= 5+10+10+5+1
=
31
Example3: Out of 7 consonants and 4 vowels, how many
words of 3 consonants and 2 vowels can be formed?
Solution: the number of ways of selecting
(3 consonants out of 7)
and (2 vowels out of 4)
= C(7,
3) x C(4, 2)
= 7 x
6 x 5 x 4 x 3
3 x 2 x
1 2 x 1
=
210
therefore
the number of group each containing 3 consonants and 2 vowels = 210