# Question 11 Exercise 6.2

Solutions of Question 11 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.

How many numbers each lying between $10$ and $1000$ can be formed with digits $2.3,4,0,8,9$ using only once?

We will form numbers greater than $10$ and less than $1000$.

So some number will consist just two digits, and some will contain three digits.

Thus we split into two parts as:
(i) Numbers greater than $10$ but less than $100$

These numbers will consist just two digits ten digit and unit digit.

Ten digit: Event $E_1$ occurs in $m_1=5$

Unit digit: Event $E_2$ occurs in $m_2=5$.

Hence the total numbers by fundamental principle of counting greater than $10$ and less than $100$ are: $$m_1 \cdot m_2=5.5=25$$

(ii) Numbers greater than $100$ and less than $1000$

In this case cach number will consist of three digits but hundred digit place can not be occupy by the digit $0$
Hundred digit: Event $E_1$ occurs in $m_1=5$

Ten digit: Event $E_2$ occurs in $\boldsymbol{m}_2=5$

Unit digit:Event $E_3$ occurs in $m_3=4$.

Hence by fundamental principle of counting numbers greater than $100$ and less than $1000$ are $$m_1 \cdot m_2 \cdot m_3=5.5 \cdot 4=100$$ Thus the total numbers greater than $10$ and less than $1000$ are: $$100 + 25=125$$