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 ⁿC_r or n or C(n, r).

Theorem

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

                  n!
     ⁿC_r =  ———
              r! (n–r)!

Theorem

^(n–1)C_(r–1) + ^(n–1)C_r = ⁿC_r

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 ⁿC_r = 126 find n Solution: Since ⁿC_r is a positive integer, we have ⁿC^r = 126 = 63 ∗ 2 = 9∗7∗2
           9∗8∗7       9∗8∗7∗6        
       =  ———   =  ——————  
              4                 4∗6
                              9∗8∗7∗6
                        =  ——————
                              4∗3∗2∗1
                         = ⁹C₄  
                    ∴n = 9
2. If ⁿC₆ = ⁿC₈, then what is ⁿC₆ ?
   Solution:
   Since ⁿC₆ = ⁿC₈ , we have n = 6 + 8 = 14
       ∴ ¹⁴C₉ = ¹⁴C₆
            14∗13∗12∗11∗ 10
       =  —————————–    = 2002
                5∗4∗3∗2∗1
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 = ⁸C₃ ∗⁵C₂= 56∗10 = 560
   6. The number of selections related to = ⁸C₂ ∗⁵C₃ = 28∗10 = 280
   7. The number of selections related to = ⁸C₁∗⁵C₄ = 8∗5 = 40
   8. The number of selections related to = ⁸C₀∗⁵C₅ = 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
   = 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.
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?
   Answer:
   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