# Question 5 and 6 Exercise 6.2

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.

## Question 5

In how many ways can letter of the word 'Fasting' be arranged?

### Solution

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}

## Question 6

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?

### Solution

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: $$m_1 \cdot m_2 \cdot m_3 \cdot m_4=2 \cdot 3 \cdot 2=12$$

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