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 $x + 1 = 8$

$\implies$ $x = 7$

Algebra - Elements of Combinatorics

Solve the equation: $\;$ $C^{n+1}_{m+1} : C^{n+1}_{m} : C^{n+1}_{m-1} = 5 : 5 : 3$

$C^{n+1}_{m+1} : C^{n+1}_{m} : C^{n+1}_{m-1} = 5 : 5 : 3$

i.e. $\;$ $C^{n+1}_{m+1} : C^{n+1}_{m} = 5 : 5$ $\;$ and $\;$ $C^{n+1}_{m} : C^{n+1}_{m-1} = 5 : 3$

Consider $\;\;\;$ $C^{n+1}_{m+1} : C^{n+1}_{m} = 5 : 5$

i.e. $\;$ $\dfrac{\left(n + 1\right)!}{\left(m + 1\right)! \left(n + 1 - m - 1\right)!} : \dfrac{\left(n + 1\right)!}{m! \left(n + 1 - m\right)!} = 1$

i.e. $\;$ $\dfrac{\left(n + 1\right)!}{\left(m + 1\right)! \left(n - m\right)!} \times \dfrac{m! \left(n + 1 - m\right)!}{\left(n + 1\right)!} = 1$

i.e. $\;$ $\dfrac{m! \left(n + 1 - m\right) \left(n - m\right)!}{\left(m + 1\right) m! \left(n - m\right)!} = 1$

i.e. $\;$ $n + 1 - m = m + 1$

i.e. $\;$ $n = 2m$ $\;\;\; \cdots \; (1)$

Consider $\;\;\;$ $C^{n+1}_{m} : C^{n+1}_{m-1} = 5 : 3$

In view of $(1)$ this becomes

$C^{2m + 1}_{m} : C^{2m + 1}_{m - 1} = 5 : 3$

i.e. $\;$ $\dfrac{\left(2m + 1\right)!}{m! \left(2m + 1 - m\right)!} : \dfrac{\left(2m + 1\right)!}{\left(m - 1\right)! \left(2m + 1 - m + 1\right)!} = \dfrac{5}{3}$

i.e. $\;$ $\dfrac{\left(m - 1\right)! \left(m + 2\right)!}{m! \left(m + 1\right)!} = \dfrac{5}{3}$

i.e. $\;$ $\dfrac{\left(m - 1\right)! \left(m + 2\right) \left(m + 1\right)!}{m \left(m -1\right)! \left(m + 1\right)!} = \dfrac{5}{3}$

i.e. $\;$ $\dfrac{m + 2}{m} = \dfrac{5}{3}$

i.e. $\;$ $3m + 6 = 5m$

i.e. $\;$ $2m = 6$ $\implies$ $m = 3$

Substituting the value of $m$ in $(1)$ gives

$n = 2m = 6$

Algebra - Elements of Combinatorics

Solve the equation: $\;$ $C^{x + 1}_{3} : C^{x}_{4} = 6 : 5$, $\;$ $x \in N$

$C^{x + 1}_{3} : C^{x}_{4} = 6 : 5$

i.e. $\;$ $\dfrac{\left(x + 1\right)!}{3! \left(x + 1 - 3\right)!} : \dfrac{x!}{4! \left(x - 4\right)!} = 6: 5$

i.e. $\;$ $\dfrac{\left(x + 1\right)!}{3! \left(x - 2\right)!} \times \dfrac{4! \left(x - 4\right)!}{x!} = \dfrac{6}{5}$

i.e. $\;$ $\dfrac{\left(x + 1\right) x!}{3! \left(x - 2\right) \left(x - 3\right) \left(x - 4\right)!} \times \dfrac{4 \times 3! \left(x - 4\right)!}{x!} = \dfrac{6}{5}$

i.e. $\;$ $\dfrac{4 \left(x + 1\right)}{\left(x - 2\right) \left(x - 3\right)}= \dfrac{6}{5}$

i.e. $\;$ $\dfrac{2 \left(x + 1\right)}{x^2 - 5x + 6} = \dfrac{3}{5}$

i.e. $\;$ $10x + 10 = 3x^2 - 15x + 18$

i.e. $\;$ $3x^2 - 25x + 8 = 0$

i.e. $\;$ $\left(x - 8\right) \left(3x - 1\right) = 0$

i.e. $\;$ $x = 8$ $\;$ or $\;$ $x = \dfrac{1}{3}$

$\because \;$ $x \in N$ $\;\;\;$ [given]

$\implies$ $x = 8$ $\;$ is the required solution.

Algebra - Elements of Combinatorics

Solve the equation: $\;$ $C^{x+1}_{x-4} = \dfrac{7}{15} P^{x+1}_{3}$, $\;$ $x \in N$

$C^{x+1}_{x-4} = \dfrac{7}{15} P^{x+1}_{3}$

i.e. $\;$ $\dfrac{\left(x + 1\right)!}{\left(x - 4\right)! \left(x + 1 - x + 4\right)!} = \dfrac{7}{15} \times \dfrac{\left(x + 1\right)!}{\left(x + 1 - 3\right)!}$

i.e. $\;$ $\dfrac{15}{\left(x - 4\right)! \times 5!} = \dfrac{7}{\left(x - 2\right)!}$

i.e. $\;$ $\dfrac{15}{\left(x - 4\right)! \times 5 \times 4 \times 3 \times 2} = \dfrac{7}{\left(x - 2\right) \left(x - 3\right) \left(x - 4\right)!}$

i.e. $\;$ $\dfrac{1}{8} = \dfrac{7}{\left(x - 2\right) \left(x - 3\right)}$

i.e. $\;$ $\left(x - 2\right) \left(x - 3\right) = 8 \times 7$

$\implies$ $x - 2 = 8$ $\;$ or $\;$ $x - 3 = 7$

$\implies$ $x = 10$

Algebra - Elements of Combinatorics

Solve the equation: $\;$ $P^{x + 1}_{3} + C^{x + 1}_{x - 1} = 14 \left(x + 1\right)$, $\;$ $x \in N$

$P^{x + 1}_{3} + C^{x + 1}_{x - 1} = 14 \left(x + 1\right)$

i.e. $\;$ $\dfrac{\left(x + 1\right)!}{\left(x + 1 - 3\right)!} + \dfrac{\left(x + 1\right)!}{\left(x - 1\right)! \left(x + 1 - x + 1\right)!} = 14 \left(x + 1\right)$

i.e. $\;$ $\dfrac{\left(x + 1\right)!}{\left(x - 2\right)!} + \dfrac{\left(x + 1\right)!}{\left(x - 1\right)! 2!} = 14 \left(x + 1\right)$

i.e. $\;$ $\dfrac{\left(x + 1\right) x \left(x - 1\right) \left(x - 2\right)!}{\left(x - 2\right)!} + \dfrac{\left(x + 1\right) x \left(x - 1\right)!}{2 \left(x - 1\right)!} = 14\left(x + 1\right)$

i.e. $\;$ $x \left(x + 1\right) \left(x - 1\right) + \dfrac{x \left(x + 1\right)}{2} = 14\left(x + 1\right)$

i.e. $\;$ $x \left(x - 1 + \dfrac{1}{2}\right) = 14$

i.e. $\;$ $\dfrac{x \left(2x - 1\right)}{2} = 14$

i.e. $\;$ $2x^2 - x - 28 = 0$

i.e. $\;$ $2 \left(x - 4\right) \left(x + 3.5\right) = 0$

$\implies$ $x = 4$, $\;$ or $\;$ $x = -3.5$

$\because \;$ $x \in N$ $\;\;\;$ [given]

$\implies$ $x = 4$ $\;$ is the required solution.