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Q.
The number of positive numbers less than $1000$ and divisible by $5$ (no digit being repeated) is
Permutations and Combinations
Solution:
Here, the available digits are $0,1,2,3,4,5,6,7,8,9$.
The numbers can be of one, two or three digits and in each of them unit's place must have $0$ or $5$ as they must be divisible by $5$ .
The number of numbers of one digit $=1$
( $\because 5$ is the only number).
The number of numbers of two digits divisible by $5=$ number of all the numbers of two digits divisible by $5$ - number of numbers of two digits divisible by $5$ and having $0$ in ten's place $={ }^{2} P_{1} \times{ }^{9} P_{1}-1$,
($\because$ unit's place can be filled by either $0$ or $5$ in first category and only by 5 in the second category)
$=2 \times 9-1=17$.
The number of numbers of three digits divisible by $5=$ number of all the numbers of three digits divisible by $5=$ number of numbers of three digits divisible by $5$ and having $0$ in hundred's place
$={ }^{2} P_{1} \times{ }^{9} P_{2}-{ }^{8} P_{1} \times 1=2 \times 9 \times 8-8=136$
$\therefore $ Required number of numbers
$=1+17+136=154$.