Example.1
(a). In how many ways can 7 men and 7 women
be seated around a circular table so that no two men/no two women sit next to
each other ?.
(b). Suppose that at a sports dinner we
have 16 cricketers and 6 tennis players. In how many ways can we seat them at a
long table if (i) none of the tennis players is seated next to another tennis
player and (ii) all tennis players are seated together.
(c). There are 20 persons among whom are
two brothers. Find the number of ways in which we can arrange them around a
circle so that there is exactly one person between the two brothers.
(d). A tea party is arranged for 2m people
along two sides of a long table with m chairs on each side. r men wish to sit
on one particular side and s on the other. In how many ways can they be seated?
(Assume that r, s £ m.).
(e). A family consists of a grandfather, m
sons and daughters and 2n grandchildren. They are to be seated in a row for
dinner. The grandchildren wish to occupy the n seats at each end and the
grandfather refuses to have a grandchild on either side of him. In how many
ways can the family be made to sit?.
Solution
(a). First arrange
the 7 men in 6! ways. Then there are 7 places in between the men which will be
occupied by 7 women in 7!, ways. So total no. of ways = 6! ´ 7!.
(b). (i) First, 16
cricketers can be seated in 16! ways. There are 17 places in between them out
of which 6 have to be occupied by tennis players.
No. of ways = 17P6 .
\ Required no. of ways = 16! ´ 17P6 .
(ii) All tennis
players can be treated as one set. So now we have a total of 17 players. Their
no. of permutations = 17 !.
Also 6 tennis
players can be arranged in 6! ways. .
No. of ways = 17 ! ´ 6!.
(c) We can arrange
18 persons around a circle in (18 –1)! = 17! ways. Now, there are exactly 18
places where we can arrange the two brothers. Also the two brothers can be
arranged in 2! ways. Thus, the number of ways of arranging the persons subject
to the given condition is (17!) (18) (2!) = 2(18!).
(d) We can arrange r
persons on m chairs on a particular side in mPr ways and
s persons on m chairs on the other side in mPs ways. We
can arrange (2m –r –s ) persons on the remaining (2m –r –s ) chairs in 2m
–r –s P2m –r –s ways. Thus, the number of ways of arranging
the persons subject to the given conditions is (mPr) (mPs)
(2m –r –s P2m –r –s).
(e) The total number
of seats required at the table is 1 + m + 2n. The grand children can occupy the
n seats on either side of the table in (2nP2n) ways. The
grandfather can occupy a seat in m –1P1 ways. The
remaining seats can be occupied in mPm ways. Therefore,
the required numbers of ways is.
(2nP2n)
(mPm) (m –1 P1) = (2n!)
m! (m –1)
Example.2
There are five
balls of different colours and five boxes of the same colours as those of the
balls. Find the number of ways in which the balls could be placed one each in a
box, so that no ball goes to the box of its own colour. .
Solution
Let us denotes the
colours by 1, 2, 3, 4 and 5. Let Ai denotes the set of distributions
of balls in the boxes so that with ith colour goes to the box of
its own colour.
We
have n( Ai) = 4! "
i,
n(Ai
Ç Aj) = 3!
( i > j) ,
n(
Ai Ç Aj ÇAk) = 2! ( i, j, k are
distinct ), etc.
We
have n(A1È A2ÈA3ÈA4ÈA5)
=
=
5 (4!) – 5C2(3!) +5C3(2!) – 5C4
(1!) +5C5
=
Thus, the number of
ways in which no ball goes into the box of its own colour is 5! –
= 60 – 20 + 5 – 1=
44.
Example.3
Six apples and six
mangoes are to be distributed among ten boys so that each boy receives at least
one fruit. Find the number of ways in which the fruits can distributed. .
Solution
When 12 fruits are distributed subjected to
the given condition, either nine boys get one fruit each and the remaining one
boy gets three fruits or eight boys get one fruit each and the remaining two
boys get two fruit each. Three fruits to a boy can be given in the four ways.
|
Apples
|
3
|
2
|
1
|
0
|
|
Mangoes
|
0
|
1
|
2
|
3
|
After giving 3 apples to a single boy, the 3
remaining apples and 6 oranges can be distributed to the 9 boys in (9C3)
(6C6) ways.
Thus, three fruits can be given to a
particular boy in
One particular boy
can be chosen in 10C1 = 10 ways
Therefore, three
fruits can be given to a single boy in
(10C1)
(2) [9C3+9C4] = 20 (10C4)
= 4200 ways
We can give two fruits to two boys, say P
and Q in the following ways.
|
P
|
Apples
|
2
|
1
|
0
|
2
|
1
|
0
|
2
|
1
|
0
|
|
Mangoes
|
0
|
1
|
2
|
0
|
1
|
2
|
0
|
1
|
2
|
|
Q
|
Apples
|
2
|
2
|
2
|
1
|
1
|
1
|
0
|
0
|
0
|
|
Mangoes
|
0
|
0
|
0
|
1
|
1
|
1
|
2
|
2
|
2
|
The remaining eight
fruits can be distributed among eight boys in the following ways: 
+
+ 
= 
Two boys out of 10
can be chosen in 10C2 ways.
Therefore, two boys
can get two fruits each in
= 
Hence, the required
number of ways = 4200 + 22050 = 26250.
Example.4
Find the number of
natural numbers which are less than 2 ´
108 and are divisible by 3 and which can be written by means of
the digits
0, 1, 2 . .
Solution
A natural number which is
smaller than 2.108 and which can be written by means of the digits
0, 1 and 2 is of the form a1 a2 … a9 where 0 £ a1
£ 1
and 0 £ ai
£ 2
for i = 2, 3, ….,9 and where all a1, a2, …, a9
cannot be equal to zero.
We can choose a1
in two ways (0 or 1) and ai for i = 2, 3,…, 8 in three ways (0, 1 or
2). After choosing a1, a2, …. , a8 we find the
sum s = a1 + a2 + …+ a8 is of the form 3m – 2,
3m – 1 or 3m. We can now choose a9 in just one way. Infact a9
= 2, 1 or 0 depending on whether S = 3m – 2, 3m – 1 or 3m. Therefore, we can
choose the numbers in (2) (37) (1) ways. But this includes a choice
in which each of ai = 0. Thus, the required number of numbers = (2)
(37) (1) – 1 = 4373.
Example.5
In an examination, the maximum marks for
each of the three papers are 50 each. Maximum marks for the fourth paper are
100. Find the number of ways in which the candidate can score 60% marks in
aggregate.
Solution
Let the marks
scored by the candidate in four papers be x1, x2, x3
and x4
Then x1+ x2+x3
+ x4 = 150 (60% of 250 = 150)
Where 0 £ x1, x2 , x3 £ 50 , 0 £ x4 £ 100.
The number of solutions of equation (1) is
same as
The ceofficient of x150 in
(1+ x + x2 +. . + x50)3 ( 1+ x + x2
+. . .+ x100).
= The ceofficient of x150 in 
= The ceofficient of x150 in
(1- x51)3 (1 – x101) (1 – x)-4
= The ceofficient
of x150 in (1 – 3x51 + 3 x102 – x101)
(1 – x)-4
= 153C3
– 3 102C3 + 3 51C3 – 52C3
= 110551.
