## Combinations

#### Definition

As we have seen earlier, if we have to pick out any two letters from the three letters a, b, c, we can do so in three different ways: ab, ac, bc. These are called combinations.

n, r are integers. If n ≥ r, for the set of n elements, a subset of r elements is called a combination.

Therefore, a combination is an unordered selection of r elements from a set of n elements.

The number of combinations of n dissimilar things taken r at a time is denoted by nCr or n or C(n, r).

#### Theorem

To find the number of combinations of “n” dissimilar things taken “r” at a time is

n!
nCr =  ———
r! (n–r)!

Theorem

(n–1)C(r–1) + (n–1)Cr = nCr

#### Corollary

If m and n are positive integers with m ≠ n, then the total m + n distinct objects can be divided into two groups of m objects and n objects in

(m+n)!
———    ways
m! n!

#### Note

(m + n + p) objects can be divided into three groups of m objects, n objects and p objects in

(m + n + p)!
—————   ways.
m! n! p!

#### Examples

1. If nCr = 126 find n Solution: Since nCr is a positive integer, we have nCr = 126 = 63 ∗ 2 = 9∗7∗2
987       9876
=  ———   =  ——————
4                 46
9876
=  ——————
4321
= 9C4
∴n = 9
2. If nC6 = nC8, then what is nC6 ?
Solution:
Since nC6 = nC8 , we have n = 6 + 8 = 14
14C9 = 14C6
14131211 10
=  —————————–    = 2002
54321
3. In how many ways can a committee of 5 members be formed from 8 men and 5 women such that at least 2 women should be in each committee?
Solution:
If there are at least 2 women in each five-member committee, then the committees are
1. 3 men, 2 women
2. 2 men, 3 women
3. 1 man, 4 women
4. No man, 5 women
5. The number of selections related to = 8C35C2= 56∗10 = 560
6. The number of selections related to = 8C25C3 = 28∗10 = 280
7. The number of selections related to = 8C15C4 = 8∗5 = 40
8. The number of selections related to = 8C05C5 = 1∗1 = 1
9. ∴ The number of ways of selecting a five-member committee is 560 + 280 + 40 + 1 = 881.
4. There are 3 questions in the first section, 3 questions in the second section and 2 questions in third section on an exam. In how many ways can a student who appeared for the exam choose to answer any 5 questions of the paper, choosing at least 1 question from each section?
Solution:
Out of 8 questions, the student can select 5 questions in following ways,
The number of ways of selecting in 1st way
= 3C23C12C1 = 6

The number of ways of selecting in 2nd way
= 3C23C22C1 = 18

The number of ways of selecting in 3rd way
= 3C23C12C2 = 9

The number of ways of selecting in 4th way
= 3C13C22C2 = 9

The total number of ways of selection = 6 + 18 + 9 + 9 = 42.
5. Find the number of divisors of 21600.
Solution
The canonical form for 21600 is 21600 = 25∗33∗52
Among the divisors of the given number, the divisor 2 may not be there, or one 2 or two 2’s or three 2’s or four 2’s or five 2’s may be there. Therefore, five 2’s can be treated as 5 + 1 = 6 ways. Similarly, treat the divisor 3 in 4 ways, 5 in 3 ways. Hence, the required number of divisors = 6∗ 4∗3 = 72. Out of these, deleting 1 and 21600, the required number of divisors is 72 - 2 = 70.

#### Try these questions

1. Find the following:
How many ways can 5 men and 3 women be seated in a row so that two women cannot sit side by side?
Since there is no restriction on men, they can sit in a row in 5! = 120 ways
Since women cannot sit side by side they can sit in between two men or in the first place or in the last place.
X M X M X M X M X M X
If the sign ‘X’ shows the women place, there are 6 vacancies for women to sit,
In these 6 vacancies 3 women can sit in 6P3 = 120 ways
∴ The number of ways where 5 men and 3 women can sit with the
given condition is
120∗120 = 14400
2. Which of the following is the correct answer? a.    If nP6 = 30 nP4 find the value of n ?
1. n = 10
2. n = 2
3. n = 8
b.    If in a factory, for 5 jobs the members A, B, C, D, E are eligible.
Then in how many ways all jobs can be filled?
1. 25
2. 20
3. 120
c.    In how many ways can 5 people sit at a round table?
1. 24
2. 120
3. 60
d.    In how many ways 7 people can sit at a round table?
1. 5040
2. 7!
3. 720

a.   Solution:

a.  n = 10

b.   Solution:

c.   120

c.   Solution:

a.   24

d.   Solution:

c.   720