Example.1
10 distinct balls have to be put in 3 different boxes.
In how many ways can we do this if the order in which we put the balls in the
boxes is also considered.
Solution
Let the balls be B1, B2,...,B10
and the boxes be b1, b2, b3.
Let us also introduce two identical markers X(say).
Now, if we take any
permutations of 10 distinct balls and these two identical markers then these
two markers divide the balls in 3 parts (in order) which we can put in the
boxes b1, b2, b3, in the same order. So each
permutation of these 12 objects in which 10 are distinct and 2 identical give a
way of the requried arrangement. So number of required ways = No. of
permutation of these 12 objects = 
.
Example.2
Suppose a man has 5 aunts and 6 uncles and his wife has
6 aunts and 5 uncles. How many way can he call a dinner party of 3 men and 3
women so that there are exactly 3 of the man’s relatives and 3 of the wife’s ?.
Solution
Looking
at the set up, we have the following cases.
|
Man
|
|
Wife
|
|
|
|
3A
|
OU
|
3U
|
OA
|
(I)
|
|
2A
|
1U
|
1A
|
2U
|
(II)
|
|
2U
|
1A
|
2A
|
1U
|
(III)
|
|
3U
|
OA
|
3A
|
OU
|
(IV)
|
A
and U denote aunt & uncle respectively.
|
Case I
|
5C3 6C0
5C3 6C0
|
100
|
|
Case II
|
5C2 6C1
6C1 5C2
|
3600
|
|
Case III
|
6C2 5C1
6C2 5C1
|
5625
|
|
Case IV
|
6C3 5C0
6C3 5C0
|
400
|
|
|
Total
|
9725
|
Example.3
In
how many ways can the letters of the word CONCUBINE be arranged so that (a) the
C's are never together (b) C's are always together.
Solution
(a) Ignoring the C's we have the letter ONUBINE
No. of ways to arrange these is 7!/2!, since there are
two Ns here. .
Also, since the C's don't have to be together, we have
the following arrangements:
X O X N X U X B X I X N X E X.
"C" can take the place of any of the positions
marked X. Since, the number of such X positions is 8 and there are 2 C's, the
number of ways we can select them is 8C2.
Hence the total number of ways is (7!/2!) 8C2.
(b)
When the C's are together, we consider them as 1 combined unit, say CC. Thus,
we have CC, O, U, B, I, N, E, N.
Number of ways we can arrange them is 8!/2! since there
are 2 N’s again.
Example.4
A colour box has 3
red colours of different shades, 2 white colours of different shades and 7 green
colours of different shades. In how many ways can 3 colours be taken from the
box if at least one of them is red ?.
Solution
We have got 3 red colours, 2 white colours and 7 green
colours.
There are 3 ways to
take 3 colours from the box.
(1)
1red 2 other
(1) 2red 1 other
(1) 3red 0 other
(1) Number of ways to choose 1red = 3C1
Number. of ways to choose 2 others = 9C2.
(2) No. of ways to choose 2red = 3C2
.
No. of ways to choose 1 others = 9C1.
(3) No. of ways to choose 3red = 3C3
.
No. of ways to choose 0 others = 9C0.
Hence the total number of ways is
3C1 x 9C2 + 3C2
x9C1 + 3C3 x9C0 .
Example.5
Let n1 = x1
x2 x3 and n2 = y1y2 y3
be two 3 digit numbers. How many pairs of n1 and n2 can
be formed so that n1 can be subtracted from n2 without
borrowing.
Solution
Clearly n1 can be subtracted from n2
without borrowing if yi ≥ xi for i = 1,
2, 3 . Let xi = r, where r = 0 to 9 for i = 2 and 3.
and r = 1 to 9 for i = 1. .
Now as per our requirement
yi = r, r +1,…. , 9. Thus we have (10 – r) choices for yi.
.
Hence total ways of choosing yi and xi
= 
.
