**Combination**

A combination is a selection of a group of objects, taken from a larger group for which the kinds of objects selected are important, but not the order in which they are selected.

There are several ways to find the number of possible combination. One is to use reasoning. Use the fundamental counting principle and divide by the number of ways that the object can be arranged among them.

**Example**: Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.

Note: AB and BA represent the same selection.

1. All the combinations formed by a, b, c taking ab, bc, ca.

2. The only combination that can be formed of three letters a, b, c taken all at a time is abc.

3. Various groups of 2 out of four persons A, B, C, D are: AB, AC, AD, BC, BD, CD.

4. Note that ab ba are two different permutations but they represent the same combination.

**Number of Combinations:**

The number of all combinations of

*n*things, taken

*r*at a time is:

combinations

**Combinations**

Let us now assume that there is a group of 3 lawn tennis players X, Y, Z. A team consisting of 2 players is to be formed. In how many ways can we do so? Is the team of X and Y different from the team of Y and X? Here, order is not important. In fact, there are only 3 possible ways in which the team could be constructed.

These are XY, YZ and ZX.

Here, each selection is called a combination of 3 different objects taken 2 at a time. In a combination, the order is not important.

Now consider some more illustrations.

Twelve persons meet in a room and each shakes hand with all the others. How do we determine the number of hand shakes. X shaking hands with Y and Y with X will not be two different hand shakes. Here, order is not important. There will be as many hand shakes as there are combinations of 12 different things taken 2 at a time.

Seven points lie on a circle. How many chords can be drawn by joining these points pairwise? There will be as many chords as there are combinations of 7 different things taken 2 at a time.

Now, we obtain the formula for finding the number of combinations of

*n*different objects taken

*r*at a time, denoted by

^{n}C

_{r}

Suppose we have 4 different objects A, B, C and D. Taking 2 at a time, if we have to make combinations, these will be AB, AC, AD, BC, BD, CD.

Here, AB and BA are the same combination as order does not alter the combination.

This is why we have not included BA, CA, DA, CB, DB and DC in this list. There are as many as 6 combinations of 4 different objects taken 2 at a time, i.e., ^{4}C_{2}=6.

Corresponding to each combination in the list, we can arrive at 2! permutations as 2 objects in each combination can be rcarrangcd in 2! ways. Hence, the number of permutations =^{4}C_{2}×2!.

On the other hand, the number of permutations of 4 different things taken 2 at a time ^{4}P_{2} . Therefore

Now, let us suppose that we have 5 different objects A, B, C. D, E. Taking 3 at a time, if we have to make combinations, these will be ABC, ABD, ABE, BCD, BCE, CDE, ACE, ACD, ADE, BDE. Corresponding to each of these

^{5}C

_{3}combinations, there are 3! permutations, because, the three objects in each combination can be rearranged in 3! ways. Therefore, the total of permutations

^{5}C

_{3}×3!. Therefore

These examples suggest the following theorem showing relationship between permutaion and combination:

^{n}P

_{r}=

^{n}C

_{r}×

*r*!,0≤

*r*≤

*n*

Proof Corresponding to each combination of

^{n}C

_{r}we have

*r*! permutations, because

*r*objects in every combination can be rearranged in

*r*! ways.

Hence, the total number of permutations of

*n*different things taken

*r*at a time is

^{n}C

_{r}×

*r*!. On the other hand, it is

^{n}P

_{r}. Thus

^{n}P

_{r}=

^{n}C

_{r}×

*r*!,0≤

*r*≤

*n*

**Examples and Solutions on Combinations**

**Example 1:**

In an examination in paper on advanced Accounts, 10 questions are set. In how many different ways can an examinee choose 7 questions?

**Solution:**

The number of different choices is evidently equal to the number of ways in which 7 places can be filled up by 10 different things.

So the number of ways are

^{10}C

_{7}=120

**Example 2:**

How many chords can be drawn through 21 points on a circle?

**Solution:**

For drawing one chord a circle, only 2 points are required.

To know the number of chords that can be drawn through the given 21 points on a circle, the number of combinations have to be counted.

Therefore, there will be as many chords as there are combinations of 21 points taken 2 at a time.

Thus, required number of chords is

**Example 3:**

In how many ways any 3 boys of a class of 20 boys are arranged in order of their increasing height? Assume that all the boys in the class had different heights.

A. 1140 B.6 C. 6840 D. None of the above

correct: A,

**Solution:**

Number of ways in which any 3 boys can be selected =

^{20}C

_{3}=1140.

**Example 4:**

In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Solution 4-I:

There are 9 courses available out of which, 2 specific courses are compulsory for every student.

Therefore, every student has to choose 3 courses out of the remaining 7 courses. This can be

chosen in ^{7}C_{3} ways.

Thus, required number of ways of choosing the programme

Solution 4-II:

Number of courses to be choosen =5, number of compulsort courses =2.

Here, number of courses =9, number of course to be choosen =5, number of compulsory courses =2, number of optional courses =7.

Use combination because each arrangement doesn’t regard the sequence.

FIG comb

Hence, required number of ways is equal to

**Example 5:**

How many committees of 5 members can be chosen from a group of 8 persons when each committee must include 2 particular persons?

**Solution:**

We have to select 6 persons out of 3 members and this can be done in

^{6}C

_{3}ways.

Thus the numbers of committees are

**Example 6:**

In how many ways can a hockey team of 11 players be selected out of 15 players? How many of them will include a particular player?

**Solution:**

The numbers of ways are

If a particular layer is to be selected in every team is kept apart, then 10 players out of 14 are

**Example 7:**

A team of 5 basketball players including a captain need to be formed out of 13 basketball players Find the number of ways in which this job can be done?

A. 13×

^{13}C

_{5}B. 8×

^{13}C

_{5}

C. 5×

^{13}C

_{5}D. Either B or C

correct: C,

**Solution:**

A team of 5 basketball players can be selected from 13 players =

^{13}C

_{5}. Now there can be 5 different captains in each team. Hence the total number of ways =

^{13}C

_{5}×5.

**Example 8:**

In how many ways, 10 boys can group themselves into two teams of 5 each?

A. 15120 B. 30240

C. 7560 D. None of the above

correct: D, **Solution:**

In combination be valid ‘AB’ and ‘BA’ is alike, if those are arranged of ‘A’ and ‘B’.

Combination operator is made by simple pattern cases such as a case bellow.

case A: How many ways can 2 boys can be taken from 4 boys?

Solution A: The respective boys are denoted by {A,B,C,D}.

Entire ways of taking 2 boys are

{AB,AC,AD,BC,BD,CD}

Hence, the required number of ways is given by

^{4}C_{2}=6

case B: How many ways can 4 boys divide themselves into two teams of 2 each?

Solution B: The respective boys are denoted by {A,B,C,D}.

Hence, the required number of ways is given by

½×^{4}C_{2}=3

Back to the main question.

No of ways of selecting five boys =^{10}C_{5}.

Hence the required number of ways

=½×^{10}C_{5}=126

**Example 9:**

In how many ways 8 boys can be distributed in two teams of 4 each to play a game of cricket?

a. 2×^{8}C_{4} b. ½×^{8}C_{4}

c. ^{8}C_{4} d. 3×^{8}C_{4}

correct: b, **Solution:**

Four boys can be selected from 8 in ^{8}C_{4} ways. The other 4 boys will form another team. Since, in this each group of 4 is considered twice, eg. Suppose there are 8 boys 1 to 8. One time a group will be formed from 1-4 and other time a group will be formed from 5-8. Both these groups will form same two pairs of teams. Hence, the total number of ways =½×^{8}C_{4} ways.

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