We know that permutation is the arrangement of objects. When we make a selection of a team containing fixed number of players, to find the number of triangles that can be formed in a polygon, number of diagonals that can be drawn in a diagonal and for other selections which are made under particular condition we use combination. This is very much useful when we solve the probability problems. In this section let us see the definition of combination, formulae, properties and few solved examples and practice problems on combinations.

## Combination Definition

Each of the different groups or selections which can be made by taking some or all of a number of things at a time (irrespective of the order) is called a combination.
By the number of combinations of n things taken r at a time is meant the number of groups of r things which can be formed from the n things. The same is denoted by the symbol nCr or C( n, r ).

The value of nCr: Each combination consists of r different things which can be arranged among themselves in r ! ways.

Therefore, the number of arrangements for all the nCr combinations is nCr x r !. This is equal to Permutations of n different things taken r at a time.

Therefore, nCr x r ! = nPr

=> nCr = $\frac{nP_{r}}{r!}$

## Combination Formula

From the above definition, we have,

nCr = $\frac{nP_{r}}{r!}$

Since nPr = $\frac{n!}{(n-r)!}$,

we get nCr = $\frac{n!}{r!\;(n-r)}!$

Case 1: When r = 0, nC0 = $\frac{n!}{0!\;(n-0)!}$

=> nC0 = $\frac{n!}{1\;(n!)}$

=> nC0 = 1

Case 2: When r = 1, nC1 = $\frac{n!}{(n-1)!\;1!}$

= $\frac{n.(n-1)!}{(n-1)!}$

nC1 = n

Case 3: When r = n, nCn = $\frac{n!}{(n-n)!\;n!}$

= $\frac{n!}{0!\;.n!}$

nCn = 1, [ since 0! = 1 ]

### Properties of Combinations:

1. nCr = nCn-r

Example 1: 5C2 = 5C(5 - 2) = 5C3

(i. e) 5C2 = $\frac{5!}{(5-2)!\;2!}$

= $\frac{5.4.3.2.1}{3.2.1.2.1}$

= $\frac{5.4}{2}$

5C2 = 10

5C3 = $\frac{5!}{(5-3)!\;3!}$

= $\frac{5.4.3.2.1}{2.1.3.2.1}$

5C3 = $\frac{5.4}{2}$ = 10

(i.e) we see that 5C2 = 5C3

Example 2: nCr + nCr-1 = (n+1)Cr
For Example. 5C3 + 5C2 = ( 5+1 )C3
We know that 5C2 = 10 and 5C3 = 10

and 6C3 = $\frac{6.5.4}{1.2.3}$

= 4.5
= 20
Also 10 + 10 = 20
Therefore, 5C3 + 5C2 = 6C3

Example 3: nCx = nCy => x = y (or) x + y = n
For Example if nC15 = nC8 , find the value of nC21
Solution: From the above property, nC15 = nC8
=> n = 15 + 8 = 23
nC21 = 23C21
= 23C23-21
= 23C2

= $\frac{23.22}{1.2}$

= 11.23
nC21 = 253

## Combination Examples

### Solved Examples

Question 1: Evaluate 18Cr = 18Cr + 2
Solution:

By the property of combination,
we have nCx = nCy => x = y ( or ) x + y = n
18Cr = 18Cr+2 => r + r + 2 = 18
=> 2r + 2 = 18
=> 2 r = 18 - 2 = 16

=> r = $\frac{16}{2}$

r = 8

Question 2: If nPr = 720 and nCr = 120, find r.
Solution:

We know that nCr = $\frac{nP_{r}}{r!}$

We have nPr = 720 and nCr = 120

Therefore, 120 = $\frac{720}{r!}$ (Substituting in the formula)

=> r! = $\frac{720}{120}$

= 6 = 3 . 2 . 1

=> r = 3 (by inspection)

Question 3: If thee are 12 persons in a party and if each two of them shake hands with each other, how many handshakes happen in the party?
Solution:

We have 12 persons in the party.
It is to note that, when two persons shake hands, it is counted as one handshake, not two . So this is the problem on combinations.
The total number of handshakes is same as the number of ways of selecting 2 persons among 12 persons = 12 C2

= $\frac{12.11}{1.2}$

= 11.6
= 66
There will be 66 handshakes in the party.

Question 4: In how many ways can a cricket team of eleven to be chosen out of a batch of 15 players if,
a. there is no restriction on selection.
b. a particular player is always chosen.
c. a particular player is never chosen.
Solution:

There are 15 players and a team of 11 is to be selected.
a. Number of ways of selecting 11 players out of 15 is 15C11 = 15C15 - 11
= 15C4

= $\frac{15.14.13.12}{1.2.3.4}$

= 1365
There are 1365 ways of selecting the team with no restriction.

b. If a particular player is is always chosen, this means that 10 players are selected out of the remaining 14 players.

Therefore, required number of ways = 14C10 = 14C14 - 10
= 14C4

= $\frac{14.13.12.11}{1.2.3.4}$

= 1001

There are 1001 ways of selecting the team when a particular player is always chosen.

c. If a particular player is never chosen, this means that 11 players are selected out of the remaining 14 players.
Therefore, required number of ways = 14C11 = 14C14 - 11
= 14C3

= $\frac{14.13.12}{1.2.3}$

= 364
There are 364 ways of selecting the team when a particular player is never chosen.

Question 5: How many triangles can be formed by joining the vertices of a hexagon?
Solution:

There are 6 vertices.
One triangle is is formed by selecting a group of 3 vertices from given 6 vertices.
This can be done in 6C3 ways.

Therefore, Number of Triangles = 6C3 = $\frac{6.5.4}{1.2.3}$

= 5 . 4
= 20
Twenty triangles can be formed by joining the vertices of a hexagon.

Question 6: In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together.
Solution:

Since boys are to be separated.
The 5 girls can be seated in 5 ! ways.
The three boys can be seated in the blanks shown in the following.

_ G _ G _ G _ G _ G _
There are 6 blank spaces, therefore, the three boys can be seated in 6C3 x 3 ! ways.
Therefore, by the Fundamental Principle of Counting,
the total number of ways = 5 ! x 6C3 x 3 !
= 14400 ways

Therefore there are 14,400 ways of seating 5 girls and 3 boys so that no two boys are together.

## Combination Practice

### Practice Problems

Question 1: If 24Cx = 24C2x + 3 , find x
Question 2: Find the value of 12C10 .
Question 3: How many different boat parties of 8, consisting of 5 boys and 3 girls, can be made from 25 boys and 10 girls?
Question 4: There are 10 points in a plane of which 4 are collinear. How many different straight lines can be drawn by joining these points?
Question 5: Find the number of diagonals formed in a decagon.