A combination is a grouping or selection of all or part of a number of things without reference to the arrangement of the things selected. Thus, the combinations of the three letters a, b, c taken 2 at a time are ab, ac, bc. Note that ab and ba are 1 combination but 2 permutations of the letters a, b.

The symbol represents the number of combinations (selections, groups) of n different things taken r at a time. Thus denotes the number of combinations of 9 different things taken 4 at a time.

Note: The symbol C(n, r) having the same meanings as is something used.

## Difference between Permutation & Combination

In combination only a group is made and the order in which the objects are arranged is immaterial.

On the other hand, in a permutation, not only a group is formed, but also an arrangement in a definite order is considered.

Example:

- ab and ba are two different permutations, but each represents the same combination.
- abc, acb, bac, bca, cab, cba are six different permutations but each one of them represents the same combination, namely a group of three objects a, b and c.

Note: We use the word ‘arrangements’ for permutations and ‘selections’ for combinations.

## Combination Evaluation

Number of combinations of n different things taken r at a time:

Example: The number of handshakes that may be exchanged among a partly of 12 students if each student shakes hands once with each other student is:

## Complementary combination

The following formula is very useful in simplifying calculations:

(Complementary combination)

This formula indicates that the number of selections of r out of n things is the same as the number of selections of n – r out of n things. Like given in the following cases:

Example: In how many ways a hockey team of eleven can be elected from 16 players?

Solution: Total number of ways

## Combination formulas

**Formula 1:** Number of combinations of n different things taken r at a time in which p particular things will always occur is .

**Formula 2:** Number of combinations of n dissimilar things taken ‘r’ at a time in which ‘p’ particular things will never occur is

Example: In a class of 25 students, find the total number of ways to select two representatives, if a particular person will never be selected.

Solution: Total students (n) = 25

A particular students will not be selected (p) = 1,

So total number of ways =

**Formula 3: **

Proof:

Evaluate

Solution:

**Formula 4: **

**Formula 5:** The number of ways in which (m + n) things can be divided into two groups containing m & n things respectively =

Example: The number of ways of dividing 3 boys & 2 girls respectively, into groups of 3 & 2 are, as follows:

Group with 3 alphabets | Group with 2 alphabets |

ABC | DE |

ABD | CE |

ACD | BE |

BCD | AE |

ABE | DC |

ACE | BD |

BCE | AD |

ADE | BC |

BDE | AC |

CDE | AB |

Example: In how many ways we can make two groups of 8 & 3 students out of total 11 students.

Solution: Total students (m + n) = 11

So total number of ways =

**Formula 6:** If 2m things are to be divided into two groups, each containing m things, the number of ways =

Example: If we divided 4 alphabets A, B, C & D into two groups containing 2 alphabets, the numbers of ways are 3. The arrangements are shown as below:

I | II |

AB | CD |

AC | BD |

AD | BC |

**Formula 7:** The number of ways to divide n things into different groups, one containing p things, another q thing and so on = where {n = p + q + r + …}

**Formula 8:** Total number of combinations of n dissimilar things taken some or all at a time = .

Example: In a city no two persons have identical set of teeth & there is no person without a tooth. Also no person has more than 32 teeth. If we disregard the shape & size of tooth & consider only the positioning of the teeth, then find the maximum population of the city.

Solution: We have 32 places for teeth. For each place we have two choice either there is a tooth or there is no tooth. Therefore the number of ways to fill up these places is 232. As there is no person without a tooth, then the maximum population is .

**Formula 9:** Total number of combination of n things, taken some or all at a time, when p of them are alike of one kind, q of them are alike of another kind and so on is: {(p+1)(q+1)(r+1)…} – 1, where n=p+q+r+…

Example: Find the total number of combinations of 5 alphabets A, B, A, B, B, taking some or all at a time.

Solution: Here A is twice & B is thrice, so by formula, total combinations = (2+1)(3+1) – 1 = 11

The combinations are as follows:

Combinations | Total | |||

1 at a time | A | B | 2 | |

2 at a time | AA | AB | BB | 3 |

3 at a time | AAB | ABB | BBB | 3 |

4 at a time | AABB | ABBB | 2 | |

5 at a time | AABBB | 1 | ||

Total | 11 |

## Exercise

1. In how many ways can 3 women be selected out of 15 women; if one particular woman is always included and two particular women are always excluded?

- 66
- 77
- 88
- 99

2. The expression would be represented by

- C(n, n)
- C(n,2)
- C(n-2,2)
- C(n-2,0)

3. Which of the below expression would be equal to C (n,0)

- C(n, 1)
- C(n, n/2)
- C(n, n-1)
- C(n, n)

4. In how many ways can a person choose 1 or more out of 4 electrical appliances?

- 10
- 12
- 14
- 15

5. Which of the below expression would be equal to C (n, r)

- C(n, 0)
- C(n, n-r)
- C(n, n)
- C(n, n/2)

6. C(n,0)+C(n,1)+ C(n,2) + C(n,3) + … + C(n, n) equals

7. C(n,1)+ C(n,2) + C(n,3) + … + C(n, n) equals

8. In a college examination, a candidate is required to answer 6 out of 10 questions which are divided into two sections each containing 5 questions. Further the candidate is not permitted to attempt more than 4 question from either of the section. The number of ways in which he can make up a choice of 6 questions is:

- 200
- 100
- 150
- 50

9. C(2n+1,0) + C(2n+1,1) + C(2n+1, 2) + … + C(2n+1,n) equals

10.C(n,0) + C(n,2) + C(n, 4) + … equal

11. C(n,1) + C(n,3) + C(n, 5) + … equal

12. In how many ways can 12 books be divided among 3 students so that each receives 4 books?

- 36540
- 34650
- 35640
- 34560

13. C(n,1) + 2C(n,2) + 3C(n,3) + … + nC(n,n) equals

14. C(n,0)+2C(n,1)+3C(n,2) + … + (n+1)C(n,0) equals

15. Evaluate

- 73
- 74
- 75
- 76

Ivan says

In the topic “Combination Evaluation” I found that that what you show “[n(n-1)(n-2)…(n-r+a)]/n!” is wrong because [n(n-1)(n-2)…(n-r+a)] < n! so the [n(n-1)(n-2)…(n-r+a)]/n! = 1