we cannot have 0 at the hundred's place. so, the hundred's place can be filled with any of the 9 digits 1, 2, 3, 4 .... , 9.
so, there are 9 ways of filling the hundred's place.
Now, 9 digits are left including 0, so, ten's place can be filled with any of the remaining 9 digits in 9 ways. now, the unit's place can be filled which in any of the remaining 8 digits. so, th ere are 8 ways of filling the unit's place.
Hence, the total number of required numbers = 9 × 9 × 8 = 648
$\therefore$ Number of ways to appointed 3 teachers = 36 × 35 × 34 = 42840
Hence, the number of ways to appoint one principal, one vice-principal and the teacher-incharge is equal to 42840.
Total number of post-offices = 5
Since a percel can be sent to any one of the five post offices.
So, the requied number of ways = 5 × 5 × 5 × 5
= 54
= 625
Hence, total number of ways is 625.
$=\frac{1}{9!}+\frac{1}{10\times9!}+\frac{1}{11\times10\times9!}$
$=\frac{11\times10+11+1}{11\times10\times9!}$
$=\frac{110+11+1}{11!}$
$=\frac{122}{11!}$
$\text{R.H.S.}$
Hence, $\frac{1}{9!}+\frac{1}{10!}+\frac{1}{11!}= \frac{122}{11!}$
This means the first digit can be selected from the 9 digits 1, 2, 3, 4 .... , 9 So, there are 9 ways of filling the first digit of the license plates.
Now, 9 digits are Ieft including 0. So, second place can be filled with any of the remaining 9 digits in 9 ways.
The third place of the license plates can be filled with in any of the remaining 8 digits. So, there are 8 ways of filling the third place.
The fourth place of the license plates can be filled with in any of the remaining 7 digits. So, there are 7 ways at filling the fourth place.
The last place of the license plates can be filled with in any of the remaining 6 digits. So, there are 6 ways of filling the fourth place.
Hence, the total number of ways = 9 × 9 × 8 × 7 × 6 = 27216
$\therefore$ First digit can be selected from the 9 digits 1, 2, 3 ..... , 9 So, there are 9 ways at filling the first digit of the licence plates.
The repetition of digits is allowed to made a license plates number.
$\therefore$ The number of ways to fill the remaining places of the number platas = 10 × 10 × 10 × 10.
Hence, the total number of ways = 9 × 10 × 10 × 10 × 10 = 90,000.
$^\text{n}\text{P}_4= 360,$
$\Rightarrow \frac{\text{n!}}{(\text{n}-4)!}=360$]
$\Rightarrow \frac{\text{n}(\text{n}-1)(\text{n}-2)(\text{n}-3)(\text{n}-4)!=360}{(\text{n}-4)!}$ $\Rightarrow \text{n}(\text{n}-1)(\text{n}-2)(\text{n}-3)=6\times 5\times4\times3 $ $\Rightarrow \text{n}=6 \ [13 \text{ by} \text{ comapring}]$ Hence, $\text{n}=6$(n + 1)! = 90 [(n - 1)!]
⇒ (n + 1)! × n × (n - 1)! = 90 [(n - 1)!] ⇒ n (n + 1) = 90 ⇒ n2 + 10n - 9n - 90 = 0 ⇒ n2 (n + 10) - 9 (n + 10) = 0 ⇒ (n2 - 9) (n + 10) = 0 $\big[\therefore \text{n} +10 \neq 0\big]$ ⇒ n - 9 = 0 ⇒ n = 9 Hence, n = 9