Combinations correspond to the selection of things from a given set of things. Here we do not intend to arrange things. We intend to select them. We intend to select them. We denote the number of unique \(r\)-selections or combinations out of a group of \(n\) objects by \({ }^n C_r\).

Combination without repetition

Combinations are selections made by taking some or all of a number of objects, irrespective of their arrangements. The number combinations of \(n\) different things taken \(r\) at a time, denoted by \({ }^n C_r\) and it is given by (no repetition, order doesn’t matter),

\(
C(n, r)={ }^n C_r=\left(\begin{array}{l}
n \\
r
\end{array}\right)=\frac{n !}{r !(n-r) !}
\)

Example 1: Out of a group of 5 people, a pair needs to be formed. Calculate the number of possible combinations.

Solution: 

\(
{ }^5 \mathrm{C}_2=\frac{5 !}{2 !(5-2) !}=\frac{5 !}{2 ! 3 !}=\frac{120}{2 \times 6}=10
\)

Example 2: Find the number of 4-letter combinations which can be made from the letters of the word DRIVEN.

Solution:

\(
{ }^6 \mathrm{C}_4=\frac{6 !}{4 !(6-4) !}=\frac{6 !}{4 ! 2 !}=\frac{720}{24 \times 2}=15
\)

Combination with repetition

\(
\frac{(r+n-1) !}{r !(n-1) !}
\)
where \({n}\) is the number of things to choose from, and we choose \({r}\) of them repetition allowed, order doesn’t matter.

Example 3: Assume there are five flavors of ice cream: banana, chocolate, lemon, strawberry, and vanilla. We can have three scoops. How many variations will there be?

Solution:

Let’s use letters for the flavors: \(\{b, c, l, s, v\}\). Example selections include

• \(\{c, c, c\}\) (3 scoops of chocolate)
• \(\{b, l, v\}\) (one each of banana, lemon and vanilla)
• \(\{b, v, v\}\) (one of banana, two of vanilla)

In this case, there are \({n}={5}\) things to choose from, we choose \({r}={3}\) of them, order does not matter, and we can repeat.

\(
\frac{(3+5-1) !}{3 !(5-1) !}=\frac{7 !}{3 ! 4 !}=\frac{5040}{6 \times 24}=35
\)
There are 35 ways of having 3 scoops from five flavors of ice cream.

Example 4:  A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?

Solution: Here, the order does not matter. Therefore, we need to count combinations. There will be as many committees as there are combinations of 5 different persons taken 3 at a time. Hence, the required number of ways \(={ }^5 \mathrm{C}_3=\frac{5 !}{3 ! 2 !}=\frac{4 \times 5}{2}=10\).
Now, 1 man can be selected from 2 men in \({ }^2 C_1\) ways and 2 women can be selected from 3 women in \({ }^3 \mathrm{C}_2\) ways. Therefore, the required number of committees

\(
={ }^2 \mathrm{C}_1 \times{ }^3 \mathrm{C}_2=\frac{2 !}{1 ! 1 !} \times \frac{3 !}{2 ! 1 !}=6
\)

Example 5: What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these
(i) four cards are of the same suit,
(ii) four cards belong to four different suits,
(iii) are face cards,
(iv) two are red cards and two are black cards,
(v) cards are of the same colour?

Solution:

There will be as many ways of choosing 4 cards from 52 cards as there are combinations of 52 different things, taken 4 at a time. Therefore
The required number of ways \(={ }^{52} \mathrm{C}_4=\frac{52 !}{4 ! 48 !}=\frac{49 \times 50 \times 51 \times 52}{2 \times 3 \times 4}\) \(=270725\)

(i) There are four suits: diamond, club, spade, heart and there are 13 cards of each suit. Therefore, there are \({ }^{13} \mathrm{C}_4\) ways of choosing 4 diamonds. Similarly, there are \({ }^{13} \mathrm{C}_4\) ways of choosing 4 clubs, \({ }^{13} \mathrm{C}_4\) ways of choosing 4 spades and \({ }^{13} \mathrm{C}_4\) ways of choosing 4 hearts. Therefore
The required number of ways \(={ }^{13} \mathrm{C}_4+{ }^{13} \mathrm{C}_4+{ }^{13} \mathrm{C}_4+{ }^{13} \mathrm{C}_4\).
\(
=4 \times \frac{13 !}{4 ! 9 !}=2860
\)

(ii) There are 13 cards in each suit.
Therefore, there are \({ }^{13} \mathrm{C}_1\) ways of choosing 1 card from 13 cards of diamond, \({ }^{13} \mathrm{C}_1\) ways of choosing 1 card from 13 cards of hearts, \({ }^{13} \mathrm{C}_1\) ways of choosing 1 card from 13 cards of clubs, \({ }^{13} \mathrm{C}_1\) ways of choosing 1 card from 13 cards of spades. Hence, by multiplication principle, the required number of ways
\(
={ }^{13} C_1 \times{ }^{13} C_1 \times{ }^{13} C_1 \times{ }^{13} C_1=13^4
\)

(iii) There are 12 face cards and 4 are to be selected out of these 12 cards. This can be done in \({ }^{12} \mathrm{C}_4\) ways. Therefore, the required number of ways \(=\frac{12 !}{4 ! 8 !}=495\).

(iv) There are 26 red cards and 26 black cards. Therefore, the required number of
\(
\begin{aligned}
\text { ways }={ }^{26} \mathrm{C}_2 & \times{ }^{26} \mathrm{C}_2 \\
&=\left(\frac{26 !}{2 ! 24 !}\right)^2=(325)^2=105625
\end{aligned}
\)

(v) 4 red cards can be selected out of 26 red cards in \({ }^{26} \mathrm{C}_4\) ways. 4 black cards can be selected out of 26 black cards in \({ }^{26} \mathrm{C}_4\) ways.
Therefore, the required number of ways \(={ }^{26} \mathrm{C}_4+{ }^{26} \mathrm{C}_4\)
\(
=2 \times \frac{26 !}{4 ! 22 !}=29900
\)

Important points about combinations

• The number of ways of selecting \(n\) objects out of \(n\) objects is(\(r=n\)):
  \(
  { }^n C_n=\frac{n !}{n !(n-n) !}=\frac{n !}{n ! 0 !}=1
  \)
• The number of ways of selecting 0 objects out of \(n\) objects is:
  \(
  { }^n C_0=\frac{n !}{0 !(n-0) !}=\frac{n !}{0 ! n !}=1
  \)
• The number of ways of selecting 1 object out of \(n\) objects is:
  \(
  { }^n C_1=\frac{n !}{1 !(n-1) !}=\frac{n \times(n-1) !}{(n-1) !}=n
  \)
• \(
  { }^n \mathrm{C}_{n-r}=\frac{n !}{(n-r) !(n-(n-r)) !}=\frac{n !}{(n-r) ! r !}={ }^n \mathrm{C}_r
  \)
  \(
  \text { i.e., selecting } r \text { objects out of } n \text { objects is same as rejecting }(n-r) \text { objects. }
  \)
• \(
  { }^n \mathrm{C}_a={ }^n \mathrm{C}_b \Rightarrow a=b \text { or } a=n-b \text {, i.e., } n=a+b
  \)
• \(
  { }^n \mathrm{C}_r+{ }^n \mathrm{C}_{r-1}={ }^{n+1} \mathrm{C}_r
  \)
• \(
  { }^n C_0+{ }^n C_1+{ }^n C_2+\ldots \ldots \ldots+{ }^n C_n=2^n
  \)
• \(
  { }^n C_0^2+{ }^n C_1^2+{ }^n C_2^2+\ldots \ldots \ldots+{ }^n C_n^2={ }^{2 n} C_n
  \)
• The relationship between permutation and combination for \(r\) things taken from \(n\) things.
  \(
  { }^n P_r=r ! \times{ }^n C_r
  \)