Posted on - 22-02-2017

IIT JEE

Meaning of combination is selection of objects.

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 = ^{n}p_{r}

&⇒ x ´ (r!) = ^{n}p_{r}.

&⇒ x =

&⇒ x = = ^{n}C_{r}
.

There are two boys B_{1} and B_{2}. B_{1}
has n_{1} different toys and B_{2} has n_{2}
different toys. Find the number of ways in which B_{1} and B_{2}
can exchange their toys in such a way that after exchanging they still
have same number of toys but not the same set. .

Total number of toys = n_{1} + n_{2}

Now let us keep all toys at one place and ask B_{1}
to pick up any n_{1} toys out of these n_{1} + n_{2}
toys . He can do it in ways
Out of these ways there is one way when he picks up those n_{1}
toys which he was initially having.

Thus required number of ways are -1.

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 ^{n}H_{r} = ^{n+r-1}C_{r} .

Let
the n objects be a_{1}, a_{2}, a_{3}… a_{n}. In
a particular group of r objects, let.

a_{1} occurs x_{1} times,

a_{2} occurs x_{2} times

a_{3} occurs x_{3} times

………………….

………………….

a_{n} occurs x_{n} times

such that x_{1} + x_{2} + x_{3}
+ ….. + x_{n} = r …. (1).

0 £ x_{i} £ 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 – 1}C_{n – 1} = ^{n + r – 1}C_{r}

Note:

Details of finding the number of integral solutions of equation (1) are given on page18.

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.

(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 3^{15} 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 ^{3}H_{15}
= ^{3 +15 - 1}C_{15} = ^{17}C_{2} ways (n = 3,
r = 5).