Combinations
A combination is a grouping or selection of all or part
of a number of things without reference to the arrangement of the
things selected. Thus, the combinations of the three letters a, b, c taken 2 at a time are ab, ac, bc. Note
that ab and ba are 1 combination but 2 permutations of the letters a, b.
The symbol nCr represents the number of combinations
(selections, groups)
of n different things taken r at a time. Thus 9C4 denotes
the number of combinations of 9 different
things taken 4 at a time.
Note: The symbol
C(n, r) having the same meanings as nCr is sometimes
used.
Difference Between Permutation & Combination
In a combination only a group is
made and the order in which the objects are arranged is immaterial.
On the other hand, in a permutation, not only a group is formed, but
also an arrangement in a definite order is considered.
Examples
1.
ab and ba are two different permutations, but each represents the same combination.
2. abc, acb, bac, bca,
cab, cba are six different permutations but
each one of them represents the same combination, namely a group of three
objects a, b and c.
things.
Remark: We
use the word 'arrangements' for
permutations and 'selections' for combinations.
Number of
combinations of n different things taken r at a time
This formula indicates that the number of selections of r out of n things is the same as the number of selections of n - r out of n
E22.In
how many ways a hockey team of eleven can be elected from 16 players?
Sol. Total
number of ways =
16C11 = 16!/11!
5!
16
x 15 x 14 x 13 x 12 / 5 x 4 x 3 x 2 x 1.
Some important formulae
Ø Number
of combinations of n different things taken r
at.a time in which p particular things will always occur is
n-pCr-p
Ø No.
of combinations of n dissimilar things taken `r' at
a time in which 'p' particular things will never occur is
n-pCr
E23.In a class of 25 students, find the total
number of ways to select two representative,
if a particular person will never be selected.
Sol. Total
students (n) = 25
A
particular students will not be selected (p) = 1,
So
total number of ways = 25-1C2 = 24C2
= 276.
E24.
Evaluate:
(i) 25C23 (ii) 75C74
Sol. (i) 25C23 = 25C25-23 = 25C2 = 300.
(ii) 75C74 = 75C75-74
= 75C1 = 75.
Ø nCr +
nCr-1 = n + 1Cr
E25. Evaluate:
(i)
5C3 + 5C2 (ii) 7C5 + 7C4
Sol. (i) 5C3 + 5C2 = 5C3 + 5C3-1 = 6C3 = 20
(ii)
7C5 + 7C4
= 7C5
+ 7C5-1 = 8C5
= 56.
Ø nC0 + nCI + nC2.....
nCn = 2n
Ø nC0 + nC2 + nC4...........
= 2n-1
Ø nC1 + nC3 + nC5...........
= 2n-1
Ø The
number of ways in which (m + n) things can be divided into two
groups containing m & n things respectively
(m+n)Cn
= (m + n)! / m! n! = (m+n)Cm
For example,
The number of ways of
dividing 3 boys & 2 girls respectively, into groups of 3 & 2
are, as follows:
Groups with 3 alphabets
|
Groups with 2 alphabets
|
ABC
|
DE
|
ABD
|
CE
|
ACD
|
BE
|
BCD
|
AE
|
ABE
|
DC
|
ACE
|
BD
|
BCE
|
AD
|
ADE
|
BC
|
BDE
|
AC
|
CDE
|
AB
|
E26.In how many ways we can
make two groups of 8 and 3 students out of total 11 students.
Sol.
Total
students (m + n) = 11
So
total number of ways = 11C8
= 165
Ø
If 2m things are to be divided into two groups, each containing m things, the
number of ways
(2m)! / 2 (m)!2
For example, if we divide 4
alphabets A, B, C and D into two groups
containing 2 alphabets, the number of ways are 3. The arrangements are
shown as below.
I
|
II
|
AB
|
CD
|
AC
|
BD
|
AD
|
BC
|
Ø The
number of ways to divide n things into different groups, one containing p
things, another q things
1
and soon = (p + q +r+...)! / (p! q! r!....) where{n
= p + q +r+...}
Ø
Total number of combinations of n dissimilar things taken some or all at a time
= 2n
- 1
E27.In a city no two
persons have identical set of teeth and there
is no person without a tooth. Also no person has more than 32 teeth.
If we disregard the shape and size of tooth and consider only the positioning
of the teeth, then find the maximum population of the city. (Assume no two
persons have similar configuration regarding positioning of teeth)
Sol. We have 32 places for
teeth. For each place we have two choices
either there is a tooth or there is no tooth. Therefore the number of
ways to fill up these places is 232. As there is no person without a tooth, then the maximum
population is 232-1.
Ø Total number of combinations of n things, taking some or all at a time, when p of them are alike of one
kind, q of them are alike of another kind and so on is
{(p+
1)(q + 1)(r + 1)...} -1, where n=p + q + r +...
E28. Find
the total number of combinations of 5 alphabets A, B, A, B, B, taking some or all at a time.
Sol. Here
A is twice and B is thrice, so by formula, total combinations = (2 + 1)
(3 + 1) — 1 = 11.
The combinations are as follows
Combinations
|
Total
|
|||
1atatime
|
A
|
B
|
2
|
|
2 at a time
|
AA
|
AB
|
BB
|
3
|
3 at a time
|
AAB
|
ABB
|
BBB
|
3
|
4 at a time
|
AABB
|
ABBB
|
2
|
|
5 at a time
|
AABBB
|
1
|
||
Total
|
11
|
|||
Miscellaneous Problems on P & C
Ø In many problems, the concepts of both
permutations and combinations are used.
Following examples will illustrate the same.
E29.Out of 6 consonants and
5 vowels, how many words of 3 consonants and 2 vowels can be formed?
Sol. Three consonants out of 6 and 2 vowels out of
5 can be chosen in
6C3 X 5C2 ways. Since each of these
words
contains 5 letters, which can be
arranged among themselves in
5! ways.
So, the
required number of words
= 6C3 X 5C2 x 5!
= 24000
E30. Eighteen guests have to be
seated, half on each side of a long table.
Four particular guests desire to sit on one particular side and three others on the other side. Determine the
number of ways in which the seating arrangement can be made.
Sol. Since
four particular guests want to sit on a particular side A (say) and three
others on the other side B (say). So, we are left with 11 guests out of which
we choose 5 for side A in 11C5
ways and the remaining 6 for side B in 6C6
ways. Hence, the number of selections for the two sides is 11C5
x 6C6. Now 9 persons on each side of the table can
be arranged among themselves in 9! ways.
Hence, the total number of arrangements
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