Algebra - Elements of Combinatorics

How many five-digit numbers, which do not contain identical digits, can be written by means of the digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ and $9$?

Given digits: $\;$ $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ and $9$ $\;$ i.e. $\;$ $9$ digits

Since the digits are not repeated,

Ten thousand's place can be selected in $9$ ways

Thousand's place can be selected in $8$ ways

Hundred's place can be selected in $7$ ways

Ten's place can be selected in $6$ ways

Unit's place can be selected in $5$ ways

$\therefore \;$ Number of five-digit numbers that can be formed with the given digits so that the digits are not repeated

$= 9 \times 8 \times 7 \times 6 \times 5 = 15120$ numbers

Algebra - Elements of Combinatorics

How many four-digit numbers which are divisible by $4$ can be formed from the digits $1$, $2$, $3$, $4$ and $5$ if the digits are repeated?

A number is divisible by $4$ if the digits in its unit's and ten's place are divisible by $4$.

Since the digits are repeated, the digits in the ten's and the unit's place can be

$1$ and $2$ $\;$ OR $\;$ $2$ and $4$ $\;$ OR $\;$ $3$ and $2$ $\;$ OR $\;$ $4$ and $4$ $\;$ OR $\;$ $5$ and $2$ $\;$ respectively

i.e. $\;$ The digits in the ten's and the unit's place can be selected in $5$ ways.

For each of these selections,

the hundred's place can be selected in $5$ ways

and the thousand's place can be selected in $5$ ways.

$\therefore \;$ Total number of four digit numbers that can be formed from the digits $1$, $2$, $3$, $4$ and $5$ (digits repeated) which are divisible by $4$ are

$5 \times 5 \times 5 = 125$ numbers

Algebra - Elements of Combinatorics

How many six-digit numbers can be formed from the digits $1$, $2$, $3$, $4$, $5$, $6$ and $7$ so that the digits do not repeat and the terminal digits should be even.

Given: $\;$ Three even digits -- $2$, $4$ and $6$

There are two terminal digits in a six digit number.

$2$ terminal digits from given $3$ even digits can be selected in

$P^3_2 = \dfrac{3!}{\left(3-2\right)!} = 3! = 6$ ways

Since the digits are not repeated, the remaining $4$ digits of the number can be selected from the remaining given $5$ digits in

$P^5_4 = 5! = 120$ ways

$\therefore \;$ Total number of six-digit numbers that can be formed from the given digits so that the digits are not repeated and the terminal digits are even are

$6 \times 120 = 720$ numbers

Algebra - Elements of Combinatorics

Five boys and three girls are sitting in a row of eight seats. In how many ways can they be seated so that not all the girls sit side by side?

Consider the three girls as $1$ group.

Then there are $5$ boys AND $1$ group of three girls, i.e. $6$ people.

These can be seated amongst themselves in $6! = 720$ ways.

Amongst themselves, the $3$ girls can be seated in $3! = 6$ ways

$\therefore \;$ Number of ways in which $5$ boys AND $1$ group of three girls can be seated

$= 3! \times 6! = 6 \times 720 = 4320$ ways.

Now, $3$ girls and $5$ boys, i.e. $8$ people can be seated amongst themselves in $8! = 40320$ ways.

$\therefore \;$ Number of ways in which $3$ girls and $5$ boys can be seated so that not all the girls sit next to one another

$= 40320 - 4320 = 36000$ ways

Algebra - Elements of Combinatorics

In how many ways can $10$ identical presents be distributed among $6$ children so that each child gets at least one present.

Let $1$ present be given to each child. This can be done in $1$ way ($\because \;$ all presents are identical).

Then we are left with $4$ identical presents to be distributed among $6$ children.

This can be done as follows:

1. Give all $4$ presents to one child.
   
   This can be done in $\;$ $C^6_1 = \dfrac{6!}{1! \times 5!} = 6$ ways
   
   OR



2. Give $3$ presents to one child and $1$ present to another.
   
   This can be done in $\;$ $C^6_1 \times C^5_1 = 6 \times 5 = 30$ ways
   
   OR



3. Give $2$ presents to one child and $2$ to another child; i.e. select $2$ children from $6$ children.
   
   This can be done in $\;$ $C^6_2 = \dfrac{6!}{2! \; 4!} = \dfrac{6 \times 5 \times 4!}{2 \times 4!} = 15$ ways
   
   OR



4. Give $2$ presents to one child and $1$ present each to two children.
   
   This can be done in $\;$ $C^6_1 \times C^5_2 = 6 \times \dfrac{5!}{2! \times 3!} = 6 \times \dfrac{5 \times 4 \times 3!}{2 \times 3!} = 60$ ways
   
   OR



5. Give $1$ present to one child each; i.e. select $4$ children from $6$ children.
   
   This can be done in $\;$ $C^6_4 = \dfrac{6!}{4! 2!} = \dfrac{6 \times 5 \times 4!}{4! \times 2} = 15$ ways

$\therefore \;$ Total number of ways in which $10$ identical presents be distributed among $6$ children is

$= 1 \times \left(6 + 30 + 15 + 60 + 15\right) = 126$ ways

Algebra - Elements of Combinatorics

In a $12$ storey house, $9$ people enter a lift cabin. It is known that they will leave the lift in groups of $2$, $3$ and $4$ people at different storeys. In how many ways can they do so if the lift does not stop at the second storey?

$9$ people can be grouped in groups of $2$, $3$ and $4$ people as follows:

$1$ group of two people $+$ $1$ group of three people $+$ $1$ group of four people $\;$ i.e. $\;$ $3$ groups

$0$ group of two people $+$ $3$ groups of three people $+$ $0$ group of four people $\;$ i.e. $\;$ $3$ groups

$3$ groups of two people $+$ $1$ group of three people $+$ $0$ group of four people $\;$ i.e. $\;$ $4$ groups

$\therefore \;$ Total number of possible groups $= 3 + 3 + 4 = 10$ groups

Now, $3$ groups can be selected from $10$ possible groups in $P^{10}_{3}$ ways.

$\therefore \;$ $9$ people entering the lift can leave in groups of $2$, $3$ and $4$ people in $P^{10}_{3}$ ways.

Algebra - Elements of Combinatorics

In how many ways can $5$ boys and $5$ girls be seated at a table so that no two boys and no two girls sit side by side?

$5$ boys can be seated at the table in $\;$ $P_5 = 5! = 120$ $\;$ ways

$5$ girls can be seated at the table in $\;$ $P_5 = 5! = 120$ $\;$ ways

Amongst themselves, the boys and girls can be seated in $\;$ $2$ $\;$ ways

$\therefore \;$ Total number of ways in which $5$ boys and $5$ girls can be seated at a table

$= 2 \times \left(P_5\right)^2 = 2 \times 120^2 = 28800$ ways

Algebra - Elements of Combinatorics

How many positive terms are there in the sequence $\left(x_n\right)$ if $x_n = \dfrac{195}{4 \times P_n} - \dfrac{P^{n+3}_{3}}{P_{n+1}}$, $\;\;\;$ $n \in N$

$\begin{aligned} x_n & = \dfrac{195}{4 \times P_n} - \dfrac{P^{n+3}_{3}}{P_{n+1}} \\\\ & = \dfrac{195}{4 \times n!} - \dfrac{\left(n+3\right)!}{\left(n+3-3\right)! \left(n+1\right)!} \\\\ & = \dfrac{195}{4 \times n!} - \dfrac{\left(n+3\right) \left(n+2\right) \left(n+1\right)!}{n! \left(n+1\right)!} \\\\ & = \dfrac{195}{4 \times n!} - \dfrac{\left(n+3\right) \left(n+2\right)}{n!} \\\\ & = \dfrac{195 - 4 \left(n^2 + 5n + 6\right)}{4 \times n!} \\\\ & = \dfrac{195 - 4n^2 - 20n - 24}{4 \times n!} \\\\ & = \dfrac{171 - 20n - 4n^2}{ 4 \times n!} \end{aligned}$

When $\;$ $n = 1$, $\;$ $x_n = \dfrac{171 - 20 - 4}{4 \times 1!} = \dfrac{147}{4 \times 1} > 0$

When $\;$ $n = 2$, $\;$ $x_n = \dfrac{171 - 40 - 16}{4 \times 2!} = \dfrac{115}{4 \times 2} > 0$

When $\;$ $n = 3$, $\;$ $x_n = \dfrac{171 - 60 - 36}{4 \times 3!} = \dfrac{75}{4 \times 6} > 0$

When $\;$ $n = 4$, $\;$ $x_n = \dfrac{171 - 80 - 64}{4 \times 4!} = \dfrac{27}{4 \times 24} > 0$

When $\;$ $n = 5$, $\;$ $x_n = \dfrac{171 - 100 - 100}{4 \times 5!} = \dfrac{-29}{4 \times 120} < 0$

$\therefore \;$ There are $4$ positive terms in the given sequence.

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $\dfrac{P^{n+1}_{4}}{C^{n-1}_{n-3}} > 14 \times P_3$, $\;$ $n \in N$

$\dfrac{P^{n+1}_{4}}{C^{n-1}_{n-3}} > 14 \times P_3$

i.e. $\;$ $\dfrac{\left(n + 1\right)!}{\left(n + 1 - 4\right)!} \div \dfrac{\left(n-1\right)!}{\left(n-3\right)! \left(n-1-n+3\right)!} > 14 \times 3!$

i.e. $\;$ $\dfrac{\left(n+1\right)!}{\left(n-3\right)!} \times \dfrac{\left(n-3\right)! \times 2!}{\left(n-1\right)!} > 14 \times 6$

i.e. $\;$ $\dfrac{2 \left(n+1\right) n \left(n-1\right)!}{\left(n-1\right)!} > 14 \times 6$

i.e. $\;$ $n \left(n+1\right) > 14 \times 3$

i.e. $\;$ $n^2 + n - 42 > 0$

i.e. $\;$ $\left(n+7\right) \left(n-6\right) > 0$

i.e. $\;$ $n+7 > 0$ $\;$ and $\;$ $n - 6 > 0$ $\;\;$ OR $\;\;$ $n + 7 < 0$ $\;$ and $\;$ $n - 6 < 0$

i.e. $\;$ $n > -7$ $\;$ and $\;$ $n > 6$ $\;\;$ OR $\;\;$ $n < -7$ $\;$ and $\;$ $n < 6$

i.e. $\;$ $n = \left\{-6, -5, -4, \cdots, 0, 1, 2, \cdots, 5, 6, 7, 8, \cdots\right\} \cap \left\{7, 8, 9, \cdots\right\}$

OR $\;\;$ $n = \left\{-8, -9, -10, \cdots\right\} \cap \left\{5, 4, 3, 2, 1, 0, -1, \cdots, -6, -7, -8, -9, \cdots\right\}$

i.e. $\;$ $n = \left\{7, 8, 9, \cdots\right\}$ $\;\;$ OR $\;\;$ $n = \left\{-8, -9, -10, \cdots\right\}$

$\because \;$ $n \in N$ $\implies$ $n = \left\{7, 8, 9, \cdots\right\}$

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $C^{x+1}_{x-1} > \dfrac{3}{2}$, $\;$ $x \in N$

$C^{x+1}_{x-1} > \dfrac{3}{2}$

i.e. $\;$ $\dfrac{\left(x+1\right)!}{\left(x-1\right)! \left(x+1-x+1\right)!} > \dfrac{3}{2}$

i.e. $\;$ $\dfrac{\left(x+1\right) x \left(x-1\right)!}{\left(x-1\right)! 2!} > \dfrac{3}{2}$

i.e. $\;$ $\dfrac{\left(x+1\right) x}{2} > \dfrac{3}{2}$

i.e. $\;$ $x^2 + x > 3$

i.e. $\;$ $x^2 + x - 3 > 0$ $\;\;\; \cdots \; (1)$

Solving the equation $\;$ $x^2 + x - 3 = 0$ $\;\;\; \cdots (2)$ gives

$x = \dfrac{-1 \pm \sqrt{1 - 4 \times 1 \times \left(-3\right)}}{2 \times 1} = \dfrac{-1 \pm \sqrt{13}}{2}$

i.e. $\;$ $x = \dfrac{-1 + \sqrt{13}}{2} = 1.302$ $\;$ or $\;$ $x = \dfrac{-1 - \sqrt{13}}{2}$

$\because \;$ $x \in N$, $\;$ the possible solution to $(1)$ is $\;$ $x \geq 2$

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $C^{x+1}_{x-1} < 21$, $\;$ $x \in N$

$C^{x+1}_{x-1} < 21$

i.e. $\;$ $\dfrac{\left(x + 1\right)!}{\left(x - 1\right)! \left(x + 1 - x + 1\right)!} < 21$

i.e. $\;$ $\dfrac{\left(x + 1\right) x \left(x - 1\right)!}{\left(x - 1\right)! 2!} < 21$

i.e. $\;$ $\left(x + 1\right) x < 42$

i.e. $\;$ $x^2 + x - 42 < 0$

i.e. $\;$ $\left(x + 7\right) \left(x - 6\right) < 0$

i.e. $\;$ $x + 7 < 0$ $\;$ and $\;$ $x - 6 > 0$ $\;\;$ OR $\;\;$ $x + 7 > 0$ $\;$ and $\;$ $x - 6 < 0$

i.e. $\;$ $x < -7$ $\;$ and $\;$ $x > 6$ $\;\;$ OR $\;\;$ $x > -7$ $\;$ and $\;$ $x < 6$

$\implies$ $x = \left\{1, 2, 3, 4, 5\right\}$ $\;$ $\because \; x \in N$

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $C^{x-1}_{4} - C^{x-1}_{3} - \dfrac{5}{4} P^{x-2}_{2} < 0$, $\;$ $x \in N$

$C^{x-1}_{4} - C^{x-1}_{3} - \dfrac{5}{4} P^{x-2}_{2} < 0$

i.e. $\;$ $C^{x-1}_{4} - C^{x-1}_{3} < \dfrac{5}{4} P^{x-2}_{2}$

i.e. $\;$ $\dfrac{\left(x - 1\right)!}{4! \left(x - 1 - 4\right)!} - \dfrac{\left(x - 1\right)!}{3! \left(x - 1 - 3\right)!} < \dfrac{5 \times \left(x - 2\right)!}{4 \times \left(x - 2 - 2\right)!}$

i.e. $\;$ $\dfrac{\left(x - 1\right)!}{4! \left(x - 5\right)!} - \dfrac{\left(x - 1\right)!}{3! \left(x - 4\right)!} < \dfrac{5 \times \left(x - 2\right)!}{4 \times \left(x - 4\right)!}$

i.e. $\;$ $\dfrac{\left(x - 1\right) \left(x - 2\right)!}{24 \left(x - 5\right)!} - \dfrac{\left(x - 1\right) \left(x - 2\right)!}{6 \left(x - 4\right) \left(x - 5\right)!} < \dfrac{5 \times \left(x - 2\right)!}{4 \left(x - 4\right) \left(x - 5\right)!}$

i.e. $\;$ $\dfrac{x-1}{12} - \dfrac{x-1}{3 \left(x - 4\right)} < \dfrac{5}{2 \left(x - 4\right)}$

i.e. $\;$ $\dfrac{\left(x - 1\right) \left(x - 4\right) - 4 \left(x - 1\right)}{12 \left(x - 4\right)} < \dfrac{5}{2 \left(x - 4\right)}$

i.e. $\;$ $\dfrac{x^2 - 5x + 4 - 4x + 4}{6} < 5$

i.e. $\;$ $x^2 - 9x + 8 < 30$

i.e. $\;$ $x^2 - 9x - 22 < 0$

i.e. $\;$ $\left(x - 11\right) \left(x + 2\right) < 0$

$\implies$ $x - 11 <0$ $\;$ and $\;$ $x + 2 > 0$ $\; \;$ OR $\;\;$ $x - 11 > 0$ $\;$ and $\;$ $x + 2 < 0$

i.e. $\;$ $x < 11$ $\;$ and $\;$ $x > -2$ $\;\;$ OR $\;\;$ $x > 11$ $\;$ and $\;$ $x < -2$

$\because \;$ $x \in N$ $\implies$ $x = \left\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\right\}$

Now, as per the question, $x - 1$ elements are taken $4$ at a time

$\implies$ $x = \left\{5, 6, 7, 8, 9,10\right\}$

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $5 C^{n}_{3} < C^{n+2}_{4}$, $\;$ $n \in N$

$5 C^{n}_{3} < C^{n+2}_{4}$

i.e. $\;$ $\dfrac{5 \times n!}{3! \left(n - 3\right)!} < \dfrac{\left(n + 2\right)!}{4! \left(n + 2 - 4\right)!}$

i.e. $\;$ $\dfrac{5 \times n!}{3! \left(n - 3\right)!} < \dfrac{\left(n + 2\right)!}{4! \left(n - 2\right)!}$

i.e. $\;$ $\dfrac{5 \times n!}{3! \left(n - 3\right)!} < \dfrac{\left(n + 2\right) \left(n + 1\right) n!}{4 \times 3! \left(n - 2\right) \left(n - 3\right)!}$

i.e. $\;$ $5 < \dfrac{\left(n + 2\right) \left(n + 1\right)}{4 \left(n - 2\right)}$

i.e. $\;$ $20n - 40 < n^2 + 3n + 2$

i.e. $\;$ $0 < n^2 - 17n + 42$

i.e. $\;$ $n^2 - 17n + 42 > 0$

i.e. $\;$ $\left(x - 14\right) \left(x - 3\right) > 0$

i.e. $\;$ $x - 14 > 0$ $\;$ or $\;$ $x - 3 > 0$

i.e. $\;$ $x > 14$ $\;$ or $\;$ $x > 3$

$\implies$ $x > 14, \; n \in N$

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $C^{n}_{6} < C^{n}_{4}$, $\;$ $n \in N$

$C^{n}_{6} < C^{n}_{4}$

i.e. $\;$ $\dfrac{n!}{6! \left(n - 6\right)!} < \dfrac{n!}{4! \left(n - 4\right)!}$

i.e. $\;$ $4! \left(n - 4\right)! < 6! \left(n - 6\right)!$

i.e. $\;$ $4! \left(n - 4\right) \left(n - 5\right) \left(n - 6\right)! < 6 \times 5 \times 4! \left(n - 6\right)!$

i.e. $\;$ $\left(n - 4\right) \left(n - 5\right) < 30$

i.e. $\;$ $n^2 -9n + 20 < 30$

i.e. $\;$ $n^2 - 9n - 10 < 0$

i.e. $\;$ $\left(n - 10\right) \left(n + 1\right) < 0$

i.e. $\;$ $n - 10 < 0$ $\;$ or $\;$ $n + 1 < 0$

i.e. $\;$ $n < 10$ $\;$ or $n < -1$

$\because \;$ $n \in N$ $\implies$ $n = \left\{6, 7, 8, 9\right\}$ $\;$ as per the question, maximum number of $n$ elements taken at a time is $6$.

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $C^{18}_{m-2} > C^{18}_{m}$, $\;$ $m \in N$

$C^{18}_{m-2} > C^{18}_{m}$

i.e. $\;$ $\dfrac{18!}{\left(m - 2\right)! \left(18 - m + 2\right)!} > \dfrac{18!}{m! \left(18 - m\right)!}$

i.e. $\;$ $m! \left(18 - m\right)! > \left(m - 2\right)! \left(20 - m\right)!$

i.e. $\;$ $m \left(m - 1\right) \left(m - 2\right)! \left(18 - m\right)! > \left(m - 2\right)! \left(20 - m\right) \left(19 - m\right) \left(18 - m\right)!$

i.e. $\;$ $m \left(m - 1\right) > \left(20 - m\right) \left(19 - m\right)$

i.e. $\;$ $m^2 - m > 380 - 39m + m^2$

i.e. $\;$ $38m > 380$

i.e. $\;$ $m > 10$

$\because \;$ $m \in M$ $\implies$ $m = \left\{11, 12, 13, 14, 15, 16, 17, 18\right\}$ $\;$ as the total number of elements is $18$.

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $C^{13}_{m} < C^{13}_{m + 2}$, $\;$ $m \in N$

$C^{13}_{m} < C^{13}_{m + 2}$

i.e. $\;$ $\dfrac{13!}{m! \left(13 - m\right)!} < \dfrac{13!}{\left(m + 2\right)! \left(13 - m - 2\right)!}$

i.e. $\;$ $\left(m + 2\right)! \left(11 - m\right)! < m! \left(13 - m\right)!$

i.e. $\;$ $\left(m + 2\right) \left(m + 1\right) m! \left(11 - m\right)! < m! \left(13 - m\right) \left(12 - m\right) \left(11 - m\right)!$

i.e. $\;$ $\left(m + 2\right) \left(m + 1\right) < \left(13 - m\right) \left(12 - m\right)$

i.e. $\;$ $m^2 + 3m + 2 < 156 - 25m + m^2$

i.e. $\;$ $28m < 154$

i.e. $\;$ $m < \dfrac{154}{28}$

i.e. $\;$ $m < \dfrac{11}{2}$

$\because \;$ $m \in N$ $\implies$ $m = \left\{1, 2, 3, 4, 5\right\}$

Algebra - Elements of Combinatorics

Solve the equation: $\;$ $3 C^{x+1}_{2} + P_2 \cdot x = 4 P^x_2$, $\;$ $x \in N$

$3 C^{x+1}_{2} + P_2 \cdot x = 4 P^x_2$

i.e. $\;$ $\dfrac{3 \times\left(x + 1\right)!}{2! \left(x + 1 -2\right)!} + 2! \times x = \dfrac{4 \times x!}{\left(x - 2\right)!}$

i.e. $\;$ $\dfrac{3 \times \left(x + 1\right)!}{2 \times \left(x - 1\right)!} + 2x = \dfrac{4 x \left(x - 1\right) \left(x - 2\right)!}{\left(x - 2\right)!}$

i.e. $\;$ $\dfrac{3 \left(x + 1\right) x \left(x - 1\right)!}{2 \left(x - 1\right)!} + 2x = 4 x \left(x - 1\right)$

i.e. $\;$ $\dfrac{3 x \left(x + 1\right)}{2} + 2x = 4x \left(x - 1\right)$

i.e. $\;$ $3x + 3 + 4 = 8x - 8$

i.e. $\;$ $5x = 15$

$\implies$ $x = 3$

Algebra - Elements of Combinatorics

Solve the equation: $\;$ $11 C^{x}_{3} = 24 C^{x+1}_{2}$, $\;\;$ $x \in N$

$11 C^{x}_{3} = 24 C^{x+1}_{2}$

i.e. $\;$ $\dfrac{11 \times x!}{3! \left(x - 3\right)!} = \dfrac{24 \times \left(x + 1\right)!}{2! \left(x + 1 - 2\right)!}$

i.e. $\;$ $\dfrac{11 \times x!}{3! \left(x - 3\right)!} = \dfrac{24 \times \left(x + 1\right)!}{2! \left(x - 1\right)!}$

i.e. $\;$ $\dfrac{11 \times x!}{ 3 \times 2! \left(x - 3\right)!} = \dfrac{24 \times \left(x + 1\right) x!}{2! \left(x - 1\right) \left(x - 2\right) \left(x - 3\right)!}$

i.e. $\;$ $\dfrac{11}{3} = \dfrac{24 \left(x + 1\right)}{\left(x - 1\right) \left(x - 2\right)}$

i.e. $\;$ $11 x^2 - 33x + 22 = 72x + 72$

i.e. $\;$ $11 x^2 - 105 x - 50 = 0$

i.e. $\;$ $\left(11x + 5\right) \left(x - 10\right) = 0$

i.e. $\;$ $11x + 5 = 0$ $\;$ or $\;$ $x - 10 = 0$

i.e. $\;$ $x = \dfrac{-5}{11}$ $\;$ or $x = 10$

$\because \;$ $x \in N$ $\implies$ $x = 10$ $\;$ is the required solution.

Algebra - Elements of Combinatorics

Solve the equation: $\;$ $C^{x}_3 + C^{x}_{4} = 11 \times C^{x+1}_{2}$, $\;\;$ $x \in N$

$C^{x}_3 + C^{x}_{4} = 11 \times C^{x+1}_{2}$

i.e. $\;$ $C^{x+1}_{4} = 11 \times C^{x+1}_{2}$ $\;\;\;$ $\left[\because \; C^{n}_{m} + C^{n}_{m + 1} = C^{n + 1}_{m + 1}\right]$

i.e. $\;$ $\dfrac{\left(x + 1\right)!}{4! \left(x + 1 - 4\right)!} = 11 \times \dfrac{\left(x + 1\right)!}{2! \left(x + 1 - 2\right)!}$

i.e. $\;$ $2! \left(x - 1\right)! = 11 \times 4! \left(x - 3\right)!$

i.e. $\;$ $2! \left(x - 1\right) \left(x - 2\right) \left(x - 3\right)! = 11 \times 4 \times 3 \times 2! \left(x - 3\right)!$

i.e. $\;$ $\left(x - 1\right) \left(x - 2\right) = 12 \times 11$

$\implies$ $x - 1 = 12$ $\;$ or $\;$ $x - 2 = 11$

$\implies$ $x = 13$

Algebra - Elements of Combinatorics

Solve the equation: $\;$ $\dfrac{P_{x + 3}}{P^{x}_{5} \times P_{x-5}} = 720$, $\;\;$ $x \in N$

$\dfrac{P_{x + 3}}{P^{x}_{5} \times P_{x-5}} = 720$

i.e. $\;$ $\dfrac{\left(x + 3\right)!}{\dfrac{x!}{\left(x - 5\right)!} \times \left(x - 5\right)!} = 720$

i.e. $\;$ $\dfrac{\left(x + 3\right)!}{x!} = 720$

i.e. $\;$ $\dfrac{\left(x + 3\right) \left(x + 2\right) \left(x + 1\right) x!}{x!} = 720$

i.e. $\;$ $\left(x + 3\right) \left(x + 2\right) \left(x + 1\right) = 10 \times 9 \times 8$

$\implies$ $x + 3 = 10$ $\;\;$ or $\;\;$ $x + 2 = 9$ $\;\;$ or $x + 1 = 8$

$\implies$ $x = 7$