Solutions of Question 7 and 8 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 three digits numbers can be formed from the digits $1,2,3,4$ and 5 if repetitions allowed?
There are three places (hundred digit, ten digit and unit digit place) to be filled by five digits,
Moreover repetition is allowed.
Hence $E_1$ occurs in $m_1=5$ ways only
$E_2$ occurs in $\cdot m_2=5$ ways
$E_3$ occurs in $m_3=5$ ways
Thus by fundamental principle of counting the total number of three digits in this case are: $$m_1 \cdot m_2 \cdot m_3=5.5 \cdot 5=125$$
How many three digits numbers can be formed from the digits $1,2,3,4$ and 5 if repetitions are not allowed?
If repetition is not allowed then each digit can appear once in each number.
In this case
$E_1$ occurs in $m_1=5$ different ways
$E_2$ occurs in $m_2=4$ ways
$E_3$ occurs in $m_3=3$ ways.
Thus by fundamental principle of 'counting the total number of three digits in this case are: $$m_1 \cdot m_2 \cdot m_3=5 \cdot 4 \cdot 3=60$$
How many different arrangements can be formed of the word “equation” if all the vowels are to be kept together?
The total number of alphabets in word equation are $8$, out of which $5$ are vowels.
If all the vowels are to kept together, then we shall deal all the vowels as a single alphabet.
So, there are four places to be filled with four alphabets.
Total places to be filled with alphabets now $=4$
Total number of ways this four places to be fillod $=4$ !
Total number of ways that 5 vowels themselves can be arrange are $=5$ !
Thus the total number of ways that the alphabets not vowel can be arrange are $=3$ !
Hence, by fundamental principle of counting the total number of different arrangements are: $4 ! \cdot 5 ! \cdot 3 !=17280$