Combinations
Meaning of
combination is selection of objects.
Selection of objects
without repetition:
The number of
selections (combinations or groups) that can be formed from n different objects
taken r (0 £ r £ n) at a time is 
Explanation:
Let the total number of selections (or groups) = x. Each
group contains r objects, which can be arranged in r! ways. Hence the number of
arrangements of r objects = x ´
(r!).
But the number of
arrangements = npr
&⇒ x ´ (r!) = npr.
&⇒ x = 
&⇒ x = 
= nCr
.
Example.1
There are two boys B1 and B2. B1
has n1 different toys and B2 has n2
different toys. Find the number of ways in which B1 and B2
can exchange their toys in such a way that after exchanging they still
have same number of toys but not the same set. .
Solution:
Total number of toys = n1 + n2
Now let us keep all toys at one place and ask B1
to pick up any n1 toys out of these n1 + n2
toys . He can do it in 
ways
Out of these ways there is one way when he picks up those n1
toys which he was initially having.
Thus required number of ways are -1.
Selection of objects with
repetition:
The number of combinations of n distinct objects, taken
r at a time when each may occur once, twice, thrice,….. upto r times in any
combination is nHr = n+r-1Cr .
Explanation:
Let
the n objects be a1, a2, a3… an. In
a particular group of r objects, let.
a1 occurs x1 times,
a2 occurs x2 times
a3 occurs x3 times
………………….
………………….
an occurs xn times
such that x1 + x2 + x3
+ ….. + xn = r …. (1).
0 £ xi £ r " i ∈
{1, 2, 3, …., n}.
Now the total number of selections of r objects out of n
= number of non-negative integral solutions of equation
(1)
= n + r – 1Cn – 1 = n + r – 1Cr
Note:
Details
of finding the number of integral solutions of equation (1) are given on
page18.
Example.2
Let 15 toys be distributed among 3 children subject to
the condition that any child can take any number of toys. Find the required
number of ways to do this if.
(i) toys are distinct. (ii) toys are
identical.
Solution:
(i) Toys are distinct
Here we have 3
children and we want the 15 toys to be distributed to the 3 children with
repetition. In other words, it is same as selecting and arranging children 15
times out of 3 children with the condition that any child can be selected any
no. of time, which can be done in 315 ways (n = 3, r = 15).
(ii)
Toys are identical
Here we only have to
select children 15 times out of 3 children with the condition that any child
can be selected any number of times which can be done in 3H15
= 3 +15 - 1C15 = 17C2 ways (n = 3,
r = 5).
