M1. Permutation and Combination:

When the order does matter, it is a Permutation. Just a collection without any regard to order or arrangement; it is a Combination.

1. Permutation:

The arrangement of objects in some order. 

a) Permutation of objects are all different: 

The total number of permutation of a set $n$, number of object taken $r$ at a time is given by:

$P(n,r)$ = $^{n}\textrm{P}_{r}$ = $\frac{n!}{(n-r)!}$ .......................................... (i)

If $r = n$; 

$P(n,n)$ = $n!$ .......................................... (ii) 

Where, $n! = 1.2.3.4 ................ (n-1).n$

Example: How many plates of vehicles consisting of $4$ different digits can be made out of the integers $4, 5, 6, 7, 8, 9$ . 

⇨ This is just like arranging $4$ objects out of $6$ integers. So we have 

     $P(6,4) = \frac{6!}{(6-4)!} = \frac{6.5.4.3.2.1}{2.1} = 360 $

b) Permutation of objects are not all different: 

The total number of permutation of a set $n$ are taken, when $p$ of the objects are of first kind, $q$ of them are of second kind, $r$ of them are of the third kind, and remaining all are different. Then,

The total number of permutations = $\frac{n!}{p!\;q!\;r!}$ .................. (iii)

Example: In how many ways can the letters of the word 'CALCULUS' be arranged? 

⇨There are $8$ letters in the word 'CALCULUS'; But $C$ comes twice, $L$ comes twice, $U$ comes twice and remaining are different.
i.e. $n = 8$; $p = 2$; $q = 2$; $r = 2$ 

Number of ways in which the letters of the given word can be arranged $ = \frac{n!}{p!\;q!\;r!}$ 

$= \frac{8!}{2!\;2!\;2!} = \frac{8*7*6*5*4*3*2*1}{2*1\;2*1\;2*1} = 5040$ 

c) Circular Permutation:
The total number of permutations of a set of $n$ objects arranged in a circle is $(n-1)!$

$P = \frac{n!}{n} = (n-1)!$ ................................ (iv)

Example: In how many ways can $7$ students be seated in a circle? 

⇨ Here, $n = 7$
The required number of ways = $(n-1)! = (7-1)! = 6! = 720$

2. Combination: 

The selection of objects without any regard to order (or, arrangement). 

The total number of combinations of $n$ objects taken $r$ at a time , Then the combination $C(n,r)$ is given as:

$^{n}\textrm{C}_{r}$ = $C(n,r)$ = $\frac{n!}{(n-r)!\;r!}$ .................................... (v)

Where, $n$ = total number of selection (combination)
             $r$ = different objects taken at a time

Example 1: A committee is to be chosen from $12$ men and $8$ women and is to consist of $3$ men and $2$ women. How many committees can be formed?

⇨ The number of ways choosing $3$ men from $12$ men is $C(12, 3) = 220$
    The number of ways choosing $2$ women from $8$ women is $C(8,2) = 28$ 

∴ Total nos of Committees formed $= 220 * 28 = 6160$

Example 2: A committee of $5$ persons is to be selected from $5$ men and $4$ ladies. In how many ways can this be done so that at least $2$ ladies are always included. 

⇨ The committee may consist of $3$ men and $2$ ladies, or $2$ men and $3$ ladies, or $1$ men and $4$ ladies. Then from these condition: 

The number of ways forming the committee consisting of $3$ men and $2$ ladies = $C(5,3) * C(4,2)$ 

= $\frac{5!}{(5-3)!\;3!} * \frac{4!}{(4-2)!\;2!} = \frac{5!}{2!\;3!} * \frac{4!}{2!\;2!} = \frac{5*4}{2} * \frac{4*3}{2} = 10 * 6 = 60\;ways$

The number of ways forming the committee consisting of $2$ men and $3$ ladies = $C(5,2) * C(4,3)$ 

= $\frac{5!}{(5-2)!\;2!} * \frac{4!}{(4-3)!\;3!} = \frac{5!}{3!\;2!} * \frac{4!}{1!\;3!} = \frac{5*4}{2} * \frac{4}{1} = 10 * 4 = 40\;ways$

The number of ways forming the committee consisting of $1$ men and $4$ ladies = $C(5,1) * C(4,4)$ 

= $\frac{5!}{(5-1)!\;1!} * \frac{4!}{(4-4)!\;4!} = \frac{5!}{4!\;1!} * \frac{4!}{0!\;4!} = \frac{5}{1} * \frac{1}{1} = 5\;ways$

∴ Total number of ways $= 60 + 40 + 5 = 105\;ways$

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M11. Probability (Mathematical terms):

"Something likely to happen !!"

1. Classical (Priori) Definition of Probability:

If $n =$ Exhaustive, Mutually exclusive and equally likely cases, and
    $m =$ Favorable cases to an event $(E)$

Then Probability of the happening of an event $(E)$ denoted by $P(E)$ is defined by,

$P(E) = \frac{m}{n}$ ................... (i)

The Probability $P(E)$ of happening of an event $E$ satisfies the following property:

$0 \leq  P(E) \leq 1$ 

Case 1. If $E$ is an impossible event then $P(E) = 0$

             If $E$ is an sure event then $P(E) = 1$

Case 2: The sum of the probabilities of the occurrence $P(E)$, and non-occurrence $P(\bar{E})$ of an event is unity. 

i.e. $P(E) + P(\bar{E}) = 1$

2. Probability:

If $S$ be the sample space of random experiment.

$n(S)$ = number of sample points of random experiment

$n(E)$ = number of favourable points to an events (E), Then, 

The Probability of happening an event $P(E)$ is defined by:

$P(E) = \frac{n(E)}{n(S)}$ ............................ (ii) 

3. Two basic laws of Probability:

i) Additional theorem:

If $A$ and $B$ are two events with their respective Probabilities $P(A)$ & $P(B)$. 

» Then the probability of occurrence of at least one of these two events denoted by $P(A \cup B)$ is given by:

$P(A$ or $B)$ = $P(A \cup B)$ = $P(A) + P(B) - P(A \cap B)$ .......(iii)

where $P(A \cap B)$ is the probability of the simultaneous occurrence of the events $A$ and $B$ (i.e common to $A$ and $B$). 

ii) Multiplication theorem:

If two events $A$ and $B$ are independent, then the probability of their simultaneous occurrence is equal to product of their individual probabilities. 

$P(A$ and $B)$ = $P(A \cap B)$ = $P(A) . P(B) $ ...........(iv)

4. Permutation and Combination:

i) Permutation: The arrangement of objects in some order.

If $n$ = total number of objects (number of permutation)

    $r$ = number of object taken (number of ways) 

» The number of permutations of a set of $(n)$, different objects taken $(r)$ at a time denoted by $P(n,r)$ or $^{n}\textrm{P}_r$ is:

$^{n}\textrm{P}_r$ = $P(n,r)$ = $\frac{n!}{(n-r)!}; (r\leq n)$ .......................... (v) 

Where $n!$ = factorial $n = 1,2,3, .............. n.$

Also,

$P(n,n)$ = $^{n}\textrm{P}_n$ = $n!$; $(0! = 1)$

» The number of permutations of a set of $n$ objects taken all of them at a time where $p$ of them are of one kind, $q$ of them the second kind, $r$ of them of the third kind. Then,

Total number of permutation = $\frac{n!}{p!q!r!}$ .......................(vi)

ii) Combination: The selection of objects without regard to any order of arrangement.

If $n$ = total number of selection (combination)
    $r$ = different objects taken at a time.

» The total number of combinations of $n$ objects taken $r$ at a time,

$C(n,r)$ = $\frac{n!}{(n-r)!r!}$ ................(vii)

5. Binomial Distribution: 

An experiment consisting only two outcomes (Success and failure) is known as "Bernoully Process". The discrete probability distribution derived from the Bernoully process is known as Binomial distribution.

Let $p$ be the probability of a success and $q$ be the probability of a failure in one trial. Let $r$ be the number of success described in $n$ independent trials of the binomial expansion of $(p+q)^n$, Then there would be $(n-r)$ failures.
The probability of getting exactly $r$ success and consequently $(n-r)$ failure in $n$ independent trials is given by;

$P(r)$ = $^{n}\textrm{C}_r \;p^r \;q^{n-r}$; $(0\leq r\leq n)$ ............... (viii)

Where, $P(r)$ = Probability of $r$ successive in $n$ trials,
                 $n$ = number of trial performed,
                 $p$ = probability of a success in a trial,
                 $q$ = probability of a failure in a trial such that $p + q = 1$
                 $r$ = number of successive in $n$ trials. (r = 0, 1, 2, ........n). 

Mean of binomial distribution = $np$ ........................ (ix)
Standard deviation of binomial distribution = $\sqrt{n\;p\;q}$ ..................(x)

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