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Basic Concepts Of Combinations for IIT and Other Engineering Exams

Posted on - 22-02-2017

JEE Math PC

IIT JEE

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).

 
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