Question 2, Exercise 6.1

Solutions of Question 2 of Exercise 6.1 of Unit 06: Permutation and Combination. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Write in the fractional form: $\quad 14\cdot 13\cdot 12\cdot 11$

Solution. \begin{align*} &14\cdot 13\cdot 12\cdot 11\\ = & \dfrac{14\cdot 13\cdot 12\cdot 11\cdot 10!}{10!} \\ = & \dfrac{14!}{10!} \end{align*}

Write in the fractional form: $\quad 1\cdot 3\cdot 5 \cdot 7 \cdot 9$

Solution.

$$1 \times 3\times 5\times 7 \times 9$$ Multiply and divided by $2\times 4\times6\times8$ \begin{align*} &\dfrac{1 \times 3\times 5\times 7\times9 \times2\times 4\times6\times8}{2\times 4\times6\times8}\\ =&\dfrac{1 \times 2\times 3\times 4\times5 \times6\times 7\times8\times9}{2^4(1\times 2\times3\times4)}\\ =& \dfrac{9!}{16 \times 4!} \\ \end{align*}

Write in the fractional form: $\quad n(n^2-1)$

Solution.

$$n(n^2-1)=n(n-1)(n+1)$$ Multiply and divided by $(n-2)!$ \begin{align*} & \dfrac{n(n-1)(n+1) \times (n-2)!}{(n-2)!} \\ & \dfrac{(n+1) n(n-1) (n-2)!}{(n-2)!} \\ =& \dfrac{(n+1)!}{(n-2)!} \end{align*}

Write in the fractional form: $\quad \dfrac{(n-3)(n-2)(n-1)}{n(n-4)}$

Solution. \begin{align*} & \dfrac{(n-3)(n-2)(n-1)}{n(n-4)} \\ = & \dfrac{(n-1)(n-2)(n-3)(n-4)!}{n(n-4)(n-4)!} \\ = & \dfrac{(n-1)!}{n(n-4)(n-4)!} \end{align*}

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