Solutions of Question 5 and 6 of Exercise 6.2 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.
In how many ways can letter of the word 'Fasting' be arranged?
The total number of alphabets in 'Fasting' are $7.$
Thus the total number of possible arrangements to fill $7$ places by these $7$ letters are: \begin{align}^7 P_7&=\dfrac{7 !}{(7-7) !}\\ & =7 !\\ &=5,040 \end{align}
How many four digits number can be formed from the digits $2,4,5,7,9$ ? (Repetition not being allowed). How many of these are even?
We have to fill four places with these five digits $2,4,5,7,9$,
so that repetition is not allowed.
We can fill the first place with 5 digits, second with the remaining 4 , third place with the remaining 3 and the fourth place with the remaining 2 digits.
Thus the total different four digits $\mathrm{n} . \mathrm{m}$ ber that can $e$ formed though these 5 digits are: $$=5.4 .3 .2=120\quad \text{or}$$
can be found using permutation as: $$^5 P_4=\dfrac{5 !}{5-4} !=120$$ Even Numbers Out of these for even number,
the unit digit have to be filled by $2$ or $4$. So, we are left with $3$ numbers.
Thus Unit digit: $E_1$ occurs in $m_1=2$
Hundred digit: $E_2$ occurs in $m_2=3$
Thousand digit: $E_3$ occurs in $m_3=2$
Ten Thousand: $E_4$ occurs in $m_4=1$.
Thus by fundamental principle of counting the total number of even numbers are: $$m_1 \cdot m_2 \cdot m_3 \cdot m_4=2 \cdot 3 \cdot 2=12$$