Permutation and Combinations - Topics, Formula, Books, FAQs

What is factorial?

A factorial is the product of all positive integers up to a given number.

It is denoted by $n!$.

$n!=n ×(n-1)×(n-2)×(n-3)×………×2 ×1$

It is calculated as follows.

$6! = 6 × 5 × 4 × 3 × 2 × 1 = 720$

Note that:

$1! = 1$

$0! = 1$

What is the meaning of $n!$?

$n!$ is read as “$n$ factorial”. It represents the products all positive integers from $1$ to $n$.

$n! = n × (n -1) × (n - 2) × (n - 3) × . . . . . . . × 1$

It is used in various mathematical and real-world contexts like:

• Permutations and combinations

• In calculations involving probability distributions.

• Algebra and calculus

• Computer science algorithms involving combinatorial logic.

What are the Fundamental Principles of Counting?

Fundamental Principle of Counting rules used to count the number of ways certain arrangements can occur.

It has two types.

• Principle of Addition

• Principle of Multiplication

Principle of Addition(OR rule)

If work $W$ can be completed by doing task $A$ OR task$ B$, and $A$ can be done in $m$ ways and $B$ can be done in $n$ ways (and both cannot occur simultaneously: in this case, we call tasks $A$ and $B$ as mutually exclusive), hen work $W$ can be done in $(m+n)$ ways.

Principle of Multiplication(AND rule)

If a certain work $W$ can be completed by doing $2$ tasks, first doing task and then doing task $B$. A can be done in $m$ ways and following that $B$ can be done in $n$ ways, then the number of ways of doing the work $W$ is $(m \times n)$ ways.

What is Permutation?

A permutation is an arrangement of all or part of a set of objects where order matters. Arranging $n$ objects in $r$ places means filling $r$ positions with $n$ distinct objects. For example, with letters $A, B, C$, the number of ways to arrange $2$ letters is $6$: $AB, AC, BA, BC, CA, CB$. Here, $n=3$ and $r=2$ show how permutations count ordered selections from $n$ objects taken $r$ at a time.

What is the formula for Permutation?

Arranging $n$ objects in $r$ places (Same as arranging $n$ objects taken $r$ at a time) is equivalent to filling $r$ places from $n$ things.

So the number of ways of arranging $n$ objects taken $r$ at a time $=n(n-1)$

$
\begin{aligned}
& (n-2) \ldots(n-r+1) \\
& \frac{n(n-1)(n-2) \ldots(n-r+1)(n-r)!}{(n-r)!}=\frac{n!}{(n-r)!}={ }^n P_r
\end{aligned}
$
Where $r \leq n$ and $r \in W$
So, the number of ways arranging n different objects taken all at a time $=$ ${ }^n P_n=n!$

Types of Permutations

There are different types of permutations.

• Permutation of $n$ different objects (when repetition is not allowed)

• Repetition, where repetition is allowed

• Permutation when the objects are not distinct (Permutation of multisets)

Permutation of $n$ different objects (when repetition is not allowed)

In this case, each object is used only once in the arrangement.

The formula to find the number of permutations of $n$ distinct objects is given by: $p(n) = n!$

Where $n!$ ($n$ factorial) is the product of all positive integers up to $n$

Permutation of $n$ different objects, where repetition is allowed

When repetition is allowed, each object can be used more than once in the arrangement. The formula to find the number of permutations when repetition is allowed is given by: $p(n,r) = nr$

Where $n$ is the total number of objects to choose from

And $r$ is the number of positions to be filled

Permutation when the objects are not distinct (Permutation of multisets)

When some objects are identical, the formula to find the number of distinct permutations is adjusted to account for the repetitions. The formula is given by:

$p(n: n_1, n_2, n_3,\ldots,......., n_k) = \frac{n!}{n_1! × n_2! × n_3! ×.........×n_k!}$

Where n is the number of total objects, $n_1, n_2, n_3,\ldots......., n_k$ are frequencies of different objects.

What is Circular Permutation?

When n objects are to be arranged circularly, then the formula to find the number of distinct permutations is given by (n-1)!.

What is the Combination?

A combination is a selection of items from a larger pool, where the order of the items does not matter. In other words, it is a way of selecting items where the arrangement or sequence of the selection is not considered.

Suppose we want to select two objects from three distinct objects a, b and c. This can be stated as a number of combinations of four different objects taken two at a time.

Here we have three different combinations $ab, bc, ca, bc$. In other words, we can say that there are three ways in which we can select two objects from three distinct objects.

Combination Formula

We can generalize this concept for $r$ object to be selected from given $n$ objects as

$
\begin{aligned}
& { }^n C_r \times r!={ }^n P_r \\
& { }^n C_r=\frac{{ }^n P_r}{r!} \\
& { }^n C_r=\frac{n!}{(n-r)!r!}
\end{aligned}
$
Where $0 \leq r \leq n$, and $r$ is a whole number.

When is the Combination to be used?

Combinations are used when the order of selection does not matter. Here are some examples of when combinations are appropriate:

1. Lottery Selection Method: In the Lottery selection method, we use the combination method. Order does not matter in selecting lotteries to choose prize winners.

2. Forming Committees: When forming a committee from a group of people, the order in which members are chosen does not matter.

3. Choosing Subsets: When choosing a subset of items from a larger set, where the order in which items are chosen is irrelevant.

4. Hand of Cards: When selecting a hand of cards from a deck, the order of the cards in the hand does not matter.

Relation between Permutation and Combination

Theorem 1: $^np_r=^nc_r × r!$

Corresponding to each combination of $^nc_r$ we have $r!$ permutations because r objects in every combination can be rearranged in $r!$ ways.

Proof:

We have to prove that $^np_r=^nc_r × r!$

Taking the left side of the equation:

$^np_r=\frac{n!}{(n-r)!}$

Taking the right side of the equation

$^nc_r × r! = \frac{n!}{(n-r!) r!} × r! = \frac{n!}{(n-r)!}=^np_r$

Hence, $^np_r=^nc_r × r!$ proved.

Theorem 2: ${ }^n\mathrm{C}_r +{ }^n\mathrm{C}_{r-1} = { }^{n+1}\mathrm{C}_r$

Proof:

We have to prove that ${ }^n\mathrm{C}_r +{ }^n\mathrm{C}_{r-1} = { }^{n+1}\mathrm{C}_r$

Taking the left side of the equation, we get,

${ }^n\mathrm{C}_r +{ }^n\mathrm{C}_{r-1}$

= $\frac{n!}{r!(n-r)!}+\frac{n!}{(n-r+1)!(r-1)!}$

= $\frac{n!}{(n-r)!r(r-1)!}+\frac{n!}{(n-r+1)!(r-1)!}$

= $\frac{n!}{(r-1)!}[\frac{1}{(n-r)!r}+\frac{1}{(n-r+1)!}]$

= $\frac{n!}{(r-1)!}[\frac{1}{r(n-r)!}+\frac{1}{(n-r+1)(n-r)!}]$

= $\frac{n!}{(r-1)!(n-r)!}[\frac{1}{r}+\frac{1}{(n-r+1)}]$

= $\frac{n!}{(r-1)!(n-r)!}[\frac{n-r+1+r}{r(n-r+1)}]$

= $\frac{n!}{(r-1)!(n-r)!}[\frac{n+1}{r(n-r+1)}]$

= $\frac{(n+1)n!}{r(r-1)!(n-r+1)(n-r)!}$

= $\frac{(n+1)!}{r!(n-r+1)!}$

= ${ }^{n+1}\mathrm{C}_r$

Hence, proved.

Difference between Permutation and Combination

To remember:

Questions based on Permutation and Combination

Question 1: The sum of the digits in the unit’s place of all the 4-digit numbers formed by using the numbers 3, 4, 5 and 6, without repetition, is :

1) 432

2) 108

3) 36

4) 18

Solution:

No. of 4 digits number formed $= 4!$

Now each digit comes equal no. of times at unit place $= 3!$

Each digit appears at the unit place

So sum $= 3! (3+4+5+6)$

$= 6 x 18 = 108$

Hence, the answer is the option 2.

Question 2: 8 - digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such numbers in which the odd digits do not occupy odd places, is :

1) 160

2) 120

3) 60

4) 48

Solution:

Odd numbers: $1, 1, 3$
Even numbers: $2, 2, 2, 4, 4$

Let us first find places for odd numbers.

$1, 1, 3$ can occupy any of the 4 even positions (2^\text{nd}, 4^\text{th}, 6^\text{th}, 8^\text{th}).
Choose 3 places for them: $\binom{4}{3} = 4$

Now, $1, 1, 3$ can be arranged in these 3 places in $\frac{3!}{2!} = 3$ ways.

Number of ways for placing odd numbers: $4 \times 3 = 12$

Now, $2, 2, 2, 4, 4$ can be arranged in the remaining 5 places.

Number of ways to arrange them: $\frac{5!}{3!2!} = 10$

So, total number of ways: $12 \times 10 = 120$

Hence, the answer is $120$.

Question 3: The number of 6-digit numbers that can be formed using the digits 0,1,2,5,7 and 9 which are divisible by 11 and no digit is repeated, is:

1) 72

2) 60

3) 48

4) 36

Solution:

The sum of the given digits is $0 + 1 + 2 + 5 + 7 + 9 = 24$.

Let the 6-digit number be $abcdef$.

It is divisible by 11 if $|(a + c + e) - (b + d + f)|$ is a multiple of $11$.

Only one possible case: $a + c + e = b + d + f = 12$.

Case 1: $(a, c, e) = (7, 5, 0)$ and $(b, d, f) = (9, 2, 1)$

So, $2! \times 2! \times 3! = 2 \times 2 \times 6 = 24$ ways.

Case 2: $(a, c, e) = (9, 2, 1)$ and $(b, d, f) = (7, 5, 0)$

So, $3! \times 3! = 6 \times 6 = 36$ ways.

Total ways: $24 + 36 = 60$

Hence, the answer is $60$.

Example 4: Using all digits from 0 to 9, how many 5-digit numbers can be formed? Repetition of digits is allowed:

1) $10^5$

2) 9 × 8 × 7 × 6 × 5

3) 90,000

4) 95

Solution:

Method 1:
We have to fill $5$ slots.
The first digit can be any from $1$ to $9$ (not $0$), so $9$ options.
Each of the remaining $4$ slots can be filled by any digit from $0$ to $9$, so $10$ options for each.
Total numbers: $9 \times 10 \times 10 \times 10 \times 10 = 90000$

Method 2:
The total ways to choose any digit for $5$ places is $10^5$.
Out of these, $10^4$ numbers start with zero (if first digit is $0$, only $4$ slots left, each with $10$ choices).
So, the count of valid five-digit numbers is $10^5 - 10^4 = 90,!000$

Hence, the answer is $90000$.

Example 5: A person had dinner at a restaurant. There were 15 South Indian and 7 Japanese food options visible on the menu card. In how many ways the person can select either a South Indian or Japanese food?

1) 7

2) 15

3) 18

4) 22
Solution:

The number of ways of selecting South Indian food $15$.

The number of ways of selecting Japanese food is $7$.

Here a person may choose any one food item, either a South Indian or a Japanese food. So, we have to use "Addition" to find the total number of ways for selecting the food item.

The total number of selecting either a South Indian or Japanese food is $15+7=22$.

Hence, the answer is the option 4.

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NCERT Resources

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NCERT Maths Class 11 Notes for Chapter 7 - Permutations and Combinations

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Practice Questions based on Permutations and Combinations

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