Example.1
Let A = {1, 2, 3,
.... ,n}. If a_{i} is the minimum element of the set A_{i}
(where A_{i} denotes the subset of A containing exactly three elements)
and X denotes the set of A_{i }s, then evaluate .
Solution
Since we can take
three elements out of n in ^{n}C_{3} ways, X will contains ^{n}C_{3}
subsets. There are exactly ^{n1}C_{2} elements of X having 1
as least element; exactly ^{n2}C_{2} elements of X having 2 as
least element and so on. In general there are exactly ^{nr}C_{2}
elements of X having r
(1 £ r £ n – 2) as least element. Thus,.
= 1(^{n1}C_{2})+2(^{n2}C_{2})
+ 3(^{n3}C_{2}) + …. + (n –2) (^{2}C_{2})
=
=
=
=
=
=
=
= .
Example.2
There are 15 seats
in a row numbered as 1 to 15. In how many ways 4 persons can sit such that seat
number 6 is always occupied in no two persons sit in adjacent seats.
Solution
Since seat number
‘6’ is always occupied remaining 3 can sit in the following ways CaseI
:
2 left to 6 and 1
right to 6. .
Two can sit in the
ways .
(1, 3) or (1, 4) or
(2, 4) and one who is sitting right to 6 can sit in the 8 ways. Hence
Total number of ways in this case = 3 ´
8 = 24.
Case–II :
1 left to 6 and 2
right to 6
The person sitting
left can sit in 4 ways
Number of ways in
which two person can sit right to 6 is same as number of integral solution of
x_{1} + x_{2} + x_{3} = 7
x_{1} ≥ 1, x_{2} ≥ 1, x_{3 }≥ 0
i.e coefficient of
t^{5} in (1 –t)^{–3} = ^{5 +3 –1}C_{3–1} = ^{7}C_{2}
= 21.
Hence total number
of ways =21´4 = 84 ways.
Case–III :
No one sits left to
6 and 3 sit right to 6. .
Number of ways.
= Number of
solution to x_{1} +x_{2} +x_{3} +x_{4} = 6(x_{1}
≥1, x_{2}
>1, x_{3} ≥ 1, x_{4}≥ 1)
= coefficient of t^{3}
in (1 –t)^{4} = ^{4+31}C_{41} = ^{6}C_{3}
= 20
Hence total number
of ways in this case = 20
Hence number of ways
of selection of seats = 24 +84 +20 =128
Total number of ways
of arrangement of 4 person on these seats
= 128 ´ 4! = 3072
Example.3
There are n straight
lines in a plane, no two of which are parallel and no three of which pass
through the same point. How many additional lines can be generated by means of
points of intersections of the given lines. .
Solution
Let AB be any one of the n straight lines
and let it be intersected by some other line CD at point E.
Note that line AB
contains (n – 1) such points. It follows that there are n (n – 1) such points
of intersection. Since each of the points of intersection occurs in two of the
n lines, the number of such points is . We shall now find the number of additional
lines passing through these points. The number of additional lines passing
through points of the type E is equal to the number of points lying outside the
line AB and CD, because a new line is obtained through E only if the other point
lies outside AB and CD. Since AB and CD each contain
(n – 2) new points, the number of new points outside AB and CD is
Thus, the number of
lines through E is . Since there are new points, the
number of new lines is
But
in this way each line is counted twice. Therefore, the required number of lines
is
=
Example.4
(a) In how many ways can 2n people be
divided into n pairs?
(b) In how many ways can you equally
distribute 100 packages of food to 10 refugees?
(c) In how many
ways mn things be equally distributed in n groups ?
(d) Find the number of ways of selecting r
pairs out of n different things.
(e) In how many ways can 22 books be divided
into 5 students so that two students will have 5 books each and the other three
students will have 4 books each.
(f) (i) In how many ways can a pack of 52
cards be divided equally among four players in order?
(ii) In how many ways can
you divide these cards into 4 sets, three of them having 17 cards each and the
fourth just one card?
Solution
(a) No. of
permutations of 2n people (2n)! Now the two people belonging to every pair can
be considered as alike and also the n pairs of same size can be considered
alike i.e. there is no arrangement of groups.
\ No. of ways =
(b) No of permutations of 100 packages =
100! The 10 packages given to each refugee can be considered as alibi. No. of
ways of making 10 groups consisting of 10 packages each =
Now distribution of packages can be done
in
(c) mn thing can be arranged in (mn)!
ways. Of these, n groups of m things each can be treated as identical and
also the m thing belong to each group can be treated as identical.
So, number of ways = .
(d) To make r pairs,
first select 2r objects out of n and then divided them into pairs.
Required number of
ways = ^{n}C_{2r} .
(e) No. of
permutations of 22 books = 22! of these, two groups of 5 books each and 3
groups of 4 books each can be treated as identical.
\ no. of ways of division into
groups =
Now these groups have to be distributed among 5 students
Required number of
ways = .
(f) (i) For the first player we have ^{52}C_{13}
choices, for the second player ^{39}C_{13} choices, for the
third player ^{26}C_{13} choices and for the last player we
have ^{13}C_{13} choices. Hence, the total number of choices.
= (^{52}C_{13}
) (^{39}C_{13} ) (^{26}C_{13} ) (^{13}C_{13}
)
= .
(ii) For the first set we
have ^{52}C_{17 }choices, for the second set ^{35}C_{17}
choices, for the third set ^{18}C_{17} choices and for the last
set we have ^{1}C_{1} choices. But the first three sets can be
interchanged in 3! ways. Hence, the total number of ways for the required
distribution.
= (^{52}C_{17}
) (^{35}C_{17} ) (^{18}C_{17} ) (^{1}C_{1}
)
Example.5
A test has 4 parts The first 3 parts
carry 10 marks each and the 4th part carries 20 marks. Assuming that marks are
not given in fractions, find the number of ways in which a candidate can get 30
marks out of 50.
Solution

Part

Max. Marks.

Marks Secured


1

10

x_{1}


2

10

x_{2}


3

10

x_{3}


4

20

x_{4}

Where
0 £x_{1}
£
10, 0 £ x_{2} £ 10
0 £ x_{3}
£
10, 0 £ x_{4} £ 20
\
No. of ways in which the candidate can get 30 marks.
=
Coefficient of x^{30} in the expression
(x°
+ x^{1} + x^{2} + …. + x^{10})^{3} (x° + x^{1}
+ x^{2} + …. + x^{20}).
=
Coefficient of x^{30} in
=
Coefficient of x^{30} in (1 – x^{11})^{3} (1 – x^{21})
(1 – x)^{4}
=
Coefficient of x^{30} in (1 – 3x^{11} + 3x^{22}) (1 – x^{21})
(1 – x)^{4}
=
Coefficient of x^{30} in (1 – 3x^{11} + 3x^{22} – x^{21})
(1 – x)^{4}
= + (3)+ (3)+ (1)
=
1111.