Question 1 and 2 Exercise 6.1

Solutions of Question 1 and 2 of Exercise 6.1 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Evaluate the $\dfrac{10 !}{3 ! .3 ! \cdot 4 !}$

$$\begin{align}\dfrac{10 !}{3 ! \cdot 3 ! \cdot 4 !}&=\dfrac{10.9 .8 \cdot 7 \cdot 6 \cdot 5.4 !}{3 ! \cdot 3 ! \cdot 4 !}\\ &=\dfrac{10.9 .8 .7 .5}{3.2 .1}\\ &=4200 \end{align}$$

Evaluate the $\dfrac{3 !+4 !}{5 !-4 !}$

$$\begin{align}\dfrac{3 !+4 !}{5 !-4 !}&=\dfrac{3 !+4.3 !}{5.4 !-4 !}\\ &=\dfrac{3 !(1+4)}{4 !(5-1)}\\ &=\dfrac{3 !(5)}{4 \cdot 3 !(4)}\\ &=\dfrac{5}{16} \end{align}$$

Evaluate the $\dfrac{(n-1) !}{(n+1) !}$

$$\begin{align}\dfrac{(n-1) !}{(n+1) !} &= \dfrac{(n-1) !}{(n+1) n(n-1) !} \\ &= \dfrac{1}{n(n+1)}\end{align}$$

Evaluate the $\dfrac{10 !}{(5 !)^2}$

$$\begin{align}\dfrac{10 !}{(5 !)^2}&=\dfrac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 !}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 5 !}\\ &=252\end{align}$$

Write $19.18.17 .16 .15 .14$ in term of factorial.

$$\begin{align}19.18.17.16.15.14& =\dfrac{19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 !}{13 !} \\ & =\dfrac{19 !}{13 !} \end{align}$$

Write $2.4.6 .8 .10 .12$ in term of factorial.

$$\begin{align}2.4,6.8 .10 .12&=2 \cdot(2 \times 2) \cdot(2 \times 3) \cdot(2 \times 2 \times 2 \times 2) \\ & (2 \times 2 \times 3) \\ & =2^6 \cdot 1 \cdot 2 \cdot 3,4 \cdot 5 \cdot 6\\ &=2^6 \cdot 6 !\end{align}$$

Write $n(n^2-1)$ in term of factorial.

$$\begin{align}n(n^2-1) &= n(n-1)(n+1) \\ & =\dfrac{(n+1) n(n-1)(n-2) !}{(n-2) !} \\ &= \dfrac{(n+1) !}{(n-2) !}\end{align}$$

Write $\dfrac{n(n+1)(n+2)}{3}$ in term of factorial.

$$\begin{align} \dfrac{n(n+1)(n+2)}{3} & =\dfrac{(n+2)(n+1) n(n-1) ! \cdot 2 !}{(n-1) ! \cdot 3 \cdot 2 !}\\ &=\dfrac{(n+2) ! \cdot 2 !}{(n-1) ! \cdot 3 !}\end{align}$$

Question 3 & 4 >