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).
To find the number of combinations of “n” dissimilar things taken “r” at a time is
nCr = ———
(n–1)C(r–1) + (n–1)Cr = nCr
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 + p) objects can be divided into three groups of m objects, n objects and p objects in
(m + n + p)!
m! n! p!
- If nCr = 126 find n
Since nCr is a positive integer, we have
nCr = 126 = 63 ∗ 2 = 9∗7∗2
= ——— = ——————
∴n = 9
If nC6 = nC8, then what is nC6 ?
Since nC6 = nC8 , we have n = 6 + 8 = 14
∴ 14C9 = 14C6
= —————————– = 2002
- 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?
If there are at least 2 women in each five-member committee, then the committees are
- 3 men, 2 women
- 2 men, 3 women
- 1 man, 4 women
- No man, 5 women
- The number of selections related to = 8C3 ∗5C2= 56∗10 = 560
- The number of selections related to = 8C2 ∗5C3 = 28∗10 = 280
- The number of selections related to = 8C1∗5C4 = 8∗5 = 40
- The number of selections related to = 8C0∗5C5 = 1∗1 = 1
- ∴ The number of ways of selecting a five-member committee is 560 + 280 + 40 + 1 = 881.
- 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?
Out of 8 questions, the student can select 5 questions in following ways,
The number of ways of selecting in 1st way
= 3C2∗3C1∗2C1 = 6
The number of ways of selecting in 2nd way
= 3C2∗3C2∗2C1 = 18
The number of ways of selecting in 3rd way
= 3C2∗3C1∗2C2 = 9
The number of ways of selecting in 4th way
= 3C1∗ 3C2∗2C2 = 9
The total number of ways of selection = 6 + 18 + 9 + 9 = 42.
- Find the number of divisors of 21600.
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
- 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
Which of the following is the correct answer?
a. If nP6 = 30 nP4 find the value of n ?
b. If in a factory, for 5 jobs the members A, B, C, D, E are eligible.
- n = 10
- n = 2
- n = 8
Then in how many ways all jobs can be filled?
c. In how many ways can 5 people sit at a round table?
d. In how many ways 7 people can sit at a round table?
a. n = 10