Formula 1: Arranging $r$ distinct objects in $r$ places = $r!$

Let there be five objects $A$, $B$, $C$, $D$, $E$ and there are five places $\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} $.

The first object $A$ takes any of the five places.

The second object $B$ takes any of the remaining four places.

....

....

The fifth object $E$ takes the last remaining place.

Total places = $5 \times 4 \times 3 \times 2 \times 1$ $ = 120$

Formula 2: Number of ways of arranging $n$ items of which $p$ are alike, $q$ alike and $t$ alike so on $ = \dfrac{{n!}}{{p! \times q! \times t! \times ...}}$

Formula 3: Selecting $r$ distinct objects out of $n$ objects $ = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

${}^n{C_0} = 1$

${}^n{C_1} = n$

${}^n{C_2} = \dfrac{{n \times \left( {n - 1} \right)}}{{2!}}$

${}^n{C_3} = \dfrac{{n \times \left( {n - 1} \right) \times \left( {n - 2} \right)}}{{3!}}$

${}^n{C_r} = {}^n{C_{n - r}}$

${}^n{C_1} = n$

${}^n{C_2} = \dfrac{{n \times \left( {n - 1} \right)}}{{2!}}$

${}^n{C_3} = \dfrac{{n \times \left( {n - 1} \right) \times \left( {n - 2} \right)}}{{3!}}$

${}^n{C_r} = {}^n{C_{n - r}}$

## Combination or selection:

In a box there are three tickets. $A,\;B,\;C$. You have to take two tickets from this box. In how many ways this can be achieved?Our possible combinations are: $AB,\; BC,\; AC$

Only 3 are possible. We don't consider the selection of $AB$ and $BA$ as two different ones. Whether you choose $A$ first and $B$ next or $B$ first and $A$ next may not make any difference.

So remember, If it is a selection like you have to select $r$ persons/ objects from $n$ persons / objects, we use ${}^n{C_{n - r}}$

Selection of 3 out of 5 is represented as ${}^5{C_3}$ which is equal to ${}^5{C_2}$ as ${}^n{C_r} = {}^n{C_{n - r}}$

How do we write ${}^5{C_3}$

Text book notation for ${}^n{C_r} = \displaystyle\frac{{n!}}{{r!(n - r)!}}$ but we hardly use this formula to calculate ${}^5{C_3}$ or any other selection.

Simply we do like this. we write 5 in the descending order upto 3 times and divide it by 3!

So ${}^5{C_3} = \displaystyle\frac{{5 \times 4 \times 3}}{{3 \times 2 \times 1}} = 10$

Or we know that ${}^n{C_r} = {}^n{C_{n - r}}$ so ${}^5{C_3} = {}^5{C_{5 - 3}} = {}^5{C_2}$

Now ${}^5{C_2} = \displaystyle\frac{{5 \times 4}}{{2 \times 1}} = 10$

Formula 4: Selecting $r$ distinct objects out of $n$ objects and arranging in $r$ places $ = {}^n{P_r} $ $= \dfrac{{n!}}{{\left( {n - r} \right)!}}$

${}^n{P_0} = 1$

${}^n{P_1} = n$

${}^n{P_2} = {n \times \left( {n - 1} \right)}$

${}^n{P_3} = {n \times \left( {n - 1} \right) \times \left( {n - 2} \right)}$

${}^n{P_n} = n!$

${}^n{P_1} = n$

${}^n{P_2} = {n \times \left( {n - 1} \right)}$

${}^n{P_3} = {n \times \left( {n - 1} \right) \times \left( {n - 2} \right)}$

${}^n{P_n} = n!$

## Permutation or Arrangement:

Now, We have to take arrange these selected letters in two places. In how many ways this can be done?Now 6 different arrangements $AB,\;$ $BA,\;$ $AC,\;$ $CA,\;$ $BC,\;$ $CB\;$ are possible.

Whenever there is an arrangement, we use the formula ${}^n{P_r}$.

So selection of $r$ objects from $n$ objects and arranging them in $r$ positions will be ${}^n{P_r}$.

Rather than using this formula, you can first select $r$ objects from $n$, using ${}^n{C_r}$ and multiplying it by $r!$ as it arranges $r$ objects in $r$ positions.

Therefore, ${}^n{P_r} = {}^n{C_{n - r}} \times r!$

Formula 5: ${}^n{P_r} = {}^n{C_r} \times r!$

Formula 6: Dividing $n$ objects where $n = p + q + r$ in such a way that first lot has $p$ objects, second lot has $q$, third lot has $r$ = $\dfrac{{n!}}{{p! \times q! \times r!}}$

Formula 7: Number of straight lines that can be formed by joining these $n$ points where no three points are collinear $ = {}^n{C_2}$

Formula 8: Number of straight lines that can be formed by joining $n$ points where $m$ points are collinear. $ = {}^n{C_2} - {}^m{C_2} + 1$

Formula 9: Number of triangles that can be formed by joining $n$ points where no three points are collinear $ = {}^n{C_3}$

Formula 10: Number of triangles that can be formed by joining $n$ points where $m$ points are collinear $ = {}^n{C_3} - {}^m{C_3}$

Formula 11: Number of triangles that can be formed by joining the vertices of a polygon of $n$ sides $ = {}^n{C_3}$

Formula 12: Number of quadrilaterals that can be formed by joining the vertices of a polygon of $n$ sides $ = {}^n{C_4}$

Formula 13: Number of rectangles that can be formed by using $m$ horizontal lines and $n$ vertical lines $ = {}^m{C_2} \times {}^n{C_2}$

Formula 14: Number of diagonals that can be formed by joining the vertices of a polygon of $n$ sides $ = \dfrac{{n\left( {n - 3} \right)}}{2}$

Formula 15: Number of ways of distributing $n$ similar articles to $r$ persons so that each gets atleast one article $ = {}^{n - 1}{C_{r - 1}}$

Formula 16: Number of ways of distributing $n$ similar articles to $r$ persons so that each gets atleast one article $ = {}^{n + r - 1}{C_{r - 1}}$