# Permutation and Combination, Binomial theorem and Probability formulas

## Permutation and Combination

Permutation and Combination types represent data from any set to a subset. In
the 17th century, the French mathematicians **Blaise Pascal** and
**Pierre de Fermat** gave the concept to develop theory of
Permutation and Combination.

### Permutation

In Permutation, we give importance to the sequence or arrangement. The permutation is the way of arranging elements of a set into subsets. Let's understand by an example.

**e.g.-** How many two-digit numbers can be formed from the set {1, 2,
3, 4}?

**Solution**-

12, 23, 34, 41

21, 32, 43, 14

11, 22, 33, 44

^{4}P_{2} = 12

Where ^{4}P_{2} = selection of two elements from
four elements in permutation.

### Combination

In combination, we choose elements from any set to form subsets without giving importance to the arrangement, order, or sequence. In other words, we don't count subsets of the same element in a different order. Let's understand by an example.

**e.g.-** Select two balls from a set of three balls(1 black, 1 white
and 1 red ball)?

**Solution**-

{Black, White}

{White, Red}

{Red, Black}

^{3}C_{2} = 3

Where ^{3}C_{2} = selection of two elements from
three elements in combination.

👉Look at this question there is no importance of arrangement, order, or
sequence **e.g.-** we can write {Black, White} or {White, Black} both are
the same.

### Difference between Permutation and Combination

Permutation | Combination |
---|---|

Order matters | The order doesn't matter |

Importance of arrangements | Importance of groups |

It mostly consists of questions related to numbers, words and symbols | In this, mostly the selection of the committee, the selection of the team, the way to solve the questions in the exam and the questions related to geometry. |

### Formula of Permutation

The number of permutations that can be made out of 'n' different things with 'r' things is-

\( ^{n}{P}_r = \frac{n!}{(n-r)!} \)

**NOTE**-

- Permutation can be represented in many ways- \( _nP_r, ^{n}{P}_r, P(n, r) \)
- \( ^{n}{P}_n = n! \)
- \( ^{n}{P}_1 = n \)
- How to solve the Factorial?

5! = 5✕4✕3✕2✕1 = 120

### Formula of Combination

The number of permutations that can be made out of 'n' different things with 'r' things is-

\( ^{n}{C}_r = \frac{n!}{r!(n-r)!} \)

**NOTE**-

- Combination can be represented in many ways- \( _nC_r, ^{n}{C}_r, C(n, r), \binom{n}{r} \)
- \( ^{n}{C}_r = \frac{1}{r!} ^{n}{P}_r \)
- \( ^{n}{C}_n = 1 \)
- \( ^{n}{C}_1 = n \)
- \( ^{n}{C}_r = ^{n}{C}_{n-r} \)

### Cyclic Permutation

When people or things are arranged in a circular order, it is used there.

**I**. When there is a difference between the objects or things arranged
clockwise and anti-clockwise = (n-1)! [It is used for Mainly living
things. ]

**II**. When there is no difference between the objects or things arranged
clockwise and anti-clockwise = \( \frac{(n-1)!}{2} \) [It is used for
Mainly non-living things. ]

Let's understand by some examples-

**e.g.I**- In how many ways can 6 persons be seated around a round table?

**Solution**-

= (n-1)!

= (6-1)!

= 5!

= 120

**e.g.II**- In how many ways can a necklace be made from 7 different types
of pearls?

**Solution**-

= \( \frac{(n-1)!}{2} \)

= \( \frac{(7-1)!}{2} \)

= \( \frac{6!}{2} \)

= \( \frac{720}{2} \)

= 360

### Some Important Points-

👉 In how many ways can one or more things be selected from 'n' different things?

2^{n} - 1

**e.g.-** In how many ways can one or more things be selected from '5'
different things?

**Solution**-

= 2^{n} - 1

= 2^{5} - 1

= 32 - 1

= 31

👉 In how many ways can one or more objects be selected from 'p' of one kind, 'q' of another and 'r' of a third kind?

(p + 1)(q + 1)(r + 1) - 1

👉 'n' things of a kind are distributed among 'r' groups or persons ―

(i) When an empty group or person is involved -

\( ^{n + r - 1}C_{r - 1} \)

**e.g.** - In how many ways can 10 identical mangoes be distributed among
four persons while any number of mangoes can be given to one person? [ This
means that all the mangoes can be given to any one person or all the mangoes
can be given to only 2 persons.]

**Solution**-

= \( ^{n + r - 1}C_{r - 1} \)

= \( ^{10 + 4 - 1}C_{4 - 1} \)

= \( ^{13}C_{3} \)

= 286

(ii) When any empty group or person is not involved -

\( ^{n - 1}C_{r - 1} \)

**e.g.** - In how many ways can 10 identical mangoes be distributed
among four persons while each person must get at least one mango?

**Solution**-

= \( ^{n - 1}C_{r - 1} \)

= \( ^{10 - 1}C_{4 - 1} \)

= \( ^{9}C_{3} \)

= 84

👉 The sum of the units digits of all the numbers formed with 'n' digits
(a_{1}, a_{2}, a_{3} ........ a_{n}) ―

(a_{1} + a_{2} + a_{3} ........
a_{n}) (n-1)!

**e.g.** - What will be the sum of the unit digits of all the 3-digit
numbers formed from the digits 2, 6, and 7?

**Solution**-

= (a_{1} + a_{2} + a_{3} ........
a_{n}) (n-1)!

= (2 + 6 + 7) (3 - 1)!

= 15 ✕ 2

= 30

👉 The sum of all the numbers formed with 'n' digits (a_{1},
a_{2}, a_{3} ........ a_{n}) ―

(a_{1} + a_{2} + a_{3} ........
a_{n}) (n-1)! (111 ..... n times)

**e.g.** - What will be the sum of all the 4 digit numbers formed from
the digits 2, 5, 6, 8?

**Solution**-

= (a_{1} + a_{2} + a_{3} ........
a_{n}) (n-1)! (111 ..... n times)

= (2 + 5 + 6 + 8) (4-1)! (1111)

= 21 ✕ 3! ✕ 1111

= 21 ✕ 6 ✕ 1111

= 139986

## Binomial theorem

You know these algebraic identities-

(x + a)^{2} = x^{2} + a^{2} + 2xa

(x + a)^{3} = x^{3} + a^{3} + 3xa(x +
a)

(x + a)^{4} = x^{4} + 4x^{3}a +
6x^{2}a^{2} + 4xa^{3} + a^{4}

You can see that as the power increases the expansion becomes more difficult. Here introduced the binomial theorem. The binomial theorem is the method that can easily expand the expression which has finite power.

### Binomial theorem for positive integers

(x +
a)^{n} = ^{n}C_{0}x^{n} a^{0} + ^{n}C_{1}x^{n-1} a^{1} + ^{n}C_{2}x^{n-2} a^{2} +
.................
+ ^{n}C_{r}x^{n-r} a^{r} +
......... + ^{n}C_{n}x^{0} a^{n}

General term
T_{r+1} = ^{n}C_{r}x^{n-r} a^{r}

### Binomial theorem for negative integers

(x -
a)^{n} = ^{n}C_{0}x^{n} a^{0} - ^{n}C_{1}x^{n-1} a^{1} + ^{n}C_{2}x^{n-2} a^{2} +
................. +
(-1)^{r} ^{n}C_{r}x^{n-r} a^{r} +
......... +
(-1)^{n} ^{n}C_{n}x^{0} a^{n}

General term T_{r+1} =
(-1)^{r} ^{n}C_{r}x^{n-r} a^{r}

**e.g.-** Expand (3x + 2)^{5}

= ^{5}C_{0}(3x)^{5} 2^{0} + ^{5}C_{1}(3x)^{5-1} 2^{1} + ^{5}C_{2}(3x)^{5-2} 2^{2} + ^{5}C_{3}(3x)^{5-3} 2^{3} + ^{5}C_{4}(3x)^{5-4} 2^{4} + ^{5}C_{5}(3x)^{5-5} 2^{5}

= 243x^{5} + 5(81x^{4}) 2 + 10(27x^{3}) 4 +
10(9x^{2}) 8 + 5(3x)16 + 32

= 243x^{5} + 810x^{4} + 1080x^{3} +
720x^{2} + 240x + 32

**e.g.-** Expand (3x - 2)^{5}

= ^{5}C_{0}(3x)^{5} 2^{0} - ^{5}C_{1}(3x)^{5-1} 2^{1} + ^{5}C_{2}(3x)^{5-2} 2^{2} - ^{5}C_{3}(3x)^{5-3} 2^{3} + ^{5}C_{4}(3x)^{5-4} 2^{4} - ^{5}C_{5}(3x)^{5-5} 2^{5}

= 243x^{5} - 5(81x^{4}) 2 + 10(27x^{3}) 4 -
10(9x^{2}) 8 + 5(3x)16 - 32

= 243x^{5} - 810x^{4} + 1080x^{3} -
720x^{2} + 240x - 32

**e.g.**- 9th term in the expansion of (2x + 2)^{13}

T_{r+1} = ^{n}C_{r}x^{n-r} a^{r}

n = 13, r = 8, x = 2x, a = 2

T_{9} =
T_{8+1} = ^{13}C_{8}(2x)^{13-8} 2^{8}

= \( \frac{13!}{8!(13-8)!} \times 32x^5 \times 256 \)

= 1287✕32x^{5}✕256

= 10543104x^{5}

### Binomial theorem for any index

### Use of Pascal's Triangle to find binomial coefficients

## Probability

Probability is the branch of mathematics it deals with random events or possibilities of occurring any events. Probability is the ratio of favourable events and Total outcomes.

\( P = \frac {Favourable \space events}{Total \space outcomes}\)

### Terminology of Probability

All terminology is given below-

#### Trail

An experiment whose result is known in advance is called a trial.

#### Event

All possible outcomes of an experiment are called events.

**e.g.-** coin toss → Trail

Heads and tails come in the toss → Events

#### Sample space

The set of all the occurrences of an experiment.

**e.g.-** Sample space of one toss of the coin → {H, T}

#### Total Outcomes

The total number of events that occurred.

#### Favourable Events

The number of occurrences of the event whose probability is to be determined.

**e.g.-** Getting the sum of the number 9 in one throw of two dice.

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

Total outcomes- 36

Favourable Events- 4 [(3, 6) (4, 5) (5, 4) (6, 3)]

👉 P(A) means the probability of happening event 'A'.

👉 P(Ā) means the probability of not happening event 'A'.

👉 P(A) + P(Ā) = 1

👉 P(A) = 1 - P(Ā)

### Mutually exclusive events

When one event occurs, the other event does not occur then the events are called mutually exclusive events.

**e.g.- **1. Suppose a dice is thrown.

The first event (E_{1}): even number coming (2, 4, 6)

The Second event (E_{2}): Odd number coming (1, 5, 3)

E_{1 }and E_{2} are mutually exclusive events.

### Addition theorem of Probability

**I**. In case of mutually exclusive events,

P(A⋃B) = P(A) + P(B)

**II**. In the case of non-mutually exclusive events,

P(A⋃B) = P(A) + P(B) - P(A⋂B)

👉 For Mutually exclusive events P(A⋂B) = 0

### Independent and Dependent Events

If the event that happened earlier has no effect on the event that happens later then the events are called independent events.

A → Toss of coin

B → Throw the dice

A & B are independent events.

E_{1} → Draw a card from a deck of cards. (52 cards)

E_{2} → Draw another card from the same deck of cards. (51 cards)

### Multiplication theorem of probability

**I**. When independent events occur,

P(AB) = P(A) P(B)

**II.** When dependent events occur,

P(AB) = P(A) P(B/A)

Where,

P(A⋃B) = P(A+B) [The probability of occurrence of event A or event B]

P(AB) = P(A⋂B) [The probability of occurrence of event A and event B]

P(B/A) = The probability of occurrence of event B when event A has already happened.

\( P(B/A) = \frac {P(A⋂B)}{P(A)} \)

### Probability distribution

Probability distribution means all possible outcomes in a trial.

#### Probability distribution table

Let's understand by example-

**e.g.**- Probability distribution of getting Head in the toss of 3 coins?

No. of Heads (X_{i}) |
0 | 1 | 2 | 3 |

Probability (P_{i}) |
1/8 | 3/8 | 3/8 | 1/8 |

**I.** Mean for Probability distribution E(x) = Σ P_{i}X_{i}

**II**. Variance for Probability distribution (σ^{2}) = Σ
P_{i}X_{i}^{2} - (Σ
P_{i}X_{i})^{2}

**III**. Standard deviation (σ) = \( \sqrt {\sum P_iX_i^2 - (\sum
P_iX_i)^2} \)

### Binomial Distribution

P_{(x = r)} = ^{n}C_{r} q^{n-r} p^{r}

Where,

n = Number of trial

r = The number of events whose probability is to be found.

P = Probability of success in an experiment.

q = 1 - p

**I.** Mean for Probability distribution E(x) = nP

**II**. Variance for Probability distribution (σ^{2}) = nPq

**III**. Standard deviation (σ) = \( \sqrt {nPq} \)

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