Combinations

What do you mean by Combination?

These two words permutation and combination, at the initial level are very confusing and are generally used interchangeably. So, let’s take them one by one and understand them.

Combination means from the given certain objects (may be alike or different) selecting one or more objects. Combination can also be replaced by the words – selection, collection or committee. 

For Example 

Combination of top 5 cricket players from the team of 11 players is the selection of 5 players (in any order).

The sequence in which they have to be selected is not important here. Also, we can say that the order of selection is not the concern in the case of combination
 

Explain Combination with different examples

Combination is the selection or collection of one or more things from the given list of alike or distinct objects taken all or some at a time.

In general, the combination of n distinct objects taken r at a time, is represented and calculated as:

This can also be represented as:

Here, in the definition of combination, r can be any positive integer less then or equals to n. 

Let’s discuss few important cases here in combination.

Theorem - III

The number of combination of n different things taken r at a time.

Example

Calculate the number of selections of 3 different colored pens from the available 5 pens of all different colors.

Here, we just need to select 3 pens (in any order) from the available 5 pens. This can be calculated as:

Here, in this theorem, note that if r = n that is, all the things have to be selected.

So 

which is very obvious. 

Theorem – IV

Number of combination of n different things taken r at a time when p particular things are always included will be calculated as 

^n-pC_r-p

Example

Calculate the number of ways of combination or selection of 11 players out of 20 players when Virat Kohli, M.S. Dhoni and Y. Singh are always included.

Here we have been giving 20 players of which only 11 players are to be selected. We are also given the 3 players out of 20 which must be included in any case. So, actually we can understand that, out of 11 we have already selected 3 players so we just need to select 8 addition players from the remaining 17 players.

Thus, Total number of ways for the above problem = ^20-3 C_11-3 = ¹⁷C₈

Theorem - V

The number of combination of n different things taken r at a time when p particular things are always excluded can be calculated as:

^n-pC_r

Let’s take the same kind of example which we have discussed in previous theorem.

Example

Calculate the number of ways of combination or selection of 11 players out of 20 players when Ravindra Jadeja and Balaji are always excluded.

Again we have been given 20 players of which 11 players to be selected but this time 2 specific players are to be excluded. Thus actually we have the option of 18 players effectively for selecting 11 players.

Thus, The total number of selections = ^20-2C₁₁ = ¹⁸C₁₁
 

How the different groups can be formed using Combination?

Let’s first understand the meaning of formation of groups. Given the objects and the number of groups to be formed, we need to select the different combination of objects to make the groups.

For Example

if we are given four players, P₁, P₂, P₃ and P₄, and if we are asked to form two groups one containing 3 players and other containing one player. Then we may form the groups as follows:

Group – 1               Group - 2

1. P₁, P₂, P₃             P₄
2. P₂, P₃, P₄             P₁
3. P₁, P₃, P₄             P₂
4. P₁, P₂, P₄             P₃

_​Thus in general, The number of ways in which (p + q) different things can be divided in two groups, one containing ‘p’ things and other contains ‘q’ things is

Also, since

^p+qC_p  = ^p+qC_q 

In the above formula, ^p+qC_p  can also be understood as selecting p (for group - 1) or rejecting ‘q’ (for group - 2) out of available (p + q) things.

Important Note:

If the size of the two groups are equal that is, p = q, then the total number of ways of formation of two groups of ‘p’ and ‘q’ things will be

Here we divide by 2! just to avoid the false counting.

To understand this false counting. Let’s again take one similar example, where we have four players P₁, P₂, P₃ and P₄ for which two groups of equal size is to be formed.

The possible combinations will be as:

Group – 1          Group - 2

1. P₁, P₂,                P₃, P₄
2. P₁, P₃,                P₂, P₄
3. P₁, P₄,                 P₂, P₃

4. P₃, P₄,                 P₁, P₂
5. P₂, P₄,                 P₁, P₃
6. P₂, P₃,                 P₁, P₄

We can clearly see that, half of the combinations are repeated. These repeated combinations are called as the false cases.

Similarly, the total number of ways in which (p+ q+ r) different things can be divided into three unequal groups p, q, and r is

Important Note:

If all the three groups are of equal size that is, (p = q = r), then the required distinct ways

Example 

In how many ways 3 teams of 11 players each, 4 teams of 6 players each, 2 teams of 15 players each can be formed out of 87 players.

Example

 In how many ways six different books can be distributed between four persons, so that each person gets at least one book.

Here we may have two cases possible – {1, 1, 1, 3} and {1, 1, 2, 2}

How the alike objects can be distributed?

Here, we have two special types for the formation of groups or distribution of alike objects.

Type – 1:

The total number of ways in which ‘n’ identical coins can be distributed to ‘p’ persons so that each person may get any number of coins is

Example

In how many ways, 8 identical coins can be distributed among 4 persons P, Q, R and S.

Here, we have 8 coins and 4 persons, so the required ways = ^8+4-1C_4-1 = ¹¹C_3.

Type – 2:

The total number of ways in which ‘n’ identical coins can be distributed to ‘p’ persons so that each person receives at least one item is

 ^n-1C_p-1

Example

In how many ways, 8 mangos can be distributed among 4 persons P, Q, R and S such that all of them receives at least one mango.

The required ways = ^8-1C_{4-1 =} ⁷C₃
 

When is ⁿC_r maximum?

If you would have studied binomial theorem, then you can easily related the ⁿC_r as the coefficient of binomial expansion. And these binomial expansions are also related with the pascal’s triangle.

In Pascal’s triangle or the binomial expansion, the value of coefficients increasing initially and reaches its maximum and then starts falling till minimum value at the end.

Thus, ⁿC_r will be maximum depending upon the value of n as r goes from 0 to n.

Thus, If n even, ⁿC_r will be maximum at r = n/2.

While, for n to be odd, ⁿC_r  will be maximum at either 

So, finally ⁿC_r is maximum at

Example

¹⁵C_r  is maximum when r = 7 or 8 while ¹⁴C_r  is maximum when r = 7
 

Is there any relation between Permutation and Combination?

As discussed in the previous sections, permutation is the combination (or selection) and then arrangement as well.

Thus, while calculating the permutation, we first need to choose or selecting the thing before their arrangement.

So, 

This can also be understood from their mathematical relation. Since we know that,

Thus, from the above two formulas, this is very clear that

ⁿP_r = ⁿC_r r!

where, ⁿC_r denotes the selection and r! denotes the arrangement of r objects for the r places.

Differentiate Permutation and Combination

The very basic difference in permutation and combination is the order of the objects considered. In combination, the order is not considered at all while for permutation it is must. So, the permutation is the ordered arrangement while the combination is the unordered selection.

From the three alphabets A, B and C, the permutation of these 3 letters will be ABC, ACB, BAC, BCA, CBA and CAB. While the combination of 3 letters will be just (A, B, C). 

Permutation gives the answer to the number of arrangements while the combination explains the possible number of selections.

Permutation of a single combination can be multiple but the combination of a single permutation is unique (considering all at a time).
 

Some Important Results

In this section, we shall take few important points with respective to the competitive examination.

Consider ‘p’ points in a plane of which q (< p) are collinear, then

• The number of different straight lines possible with these p points is

C₂ - ^qC₂ + 1

• The number of different triangles possible with these p points is

C₃ - ^qC₃

• The number of diagonals in polygon of p sides is

C₂ - p

• If ‘q’ parallel lines are intersected by the ‘p’ different set of parallel lines, then the total number of parallelograms so formed is

C₂ x ^qC₂

• The number of triangles possible in a polygon of p sides is ^pC₃

  • The number of triangles having exactly two sides common to the polygon = p

  • The number of triangles having exactly one side common to the polygon = p (p-4)

  • The number of triangles having no sides common to the polygon = p(p-4)(p-5)/6
     

Watch this Video for more reference
 

More Readings

Combinations

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