**Question:**

How many numbers divisible by 5 and lying between 3000 and 4000 can be formed by using the digits 3, 4, 5, 6, 7, 8 when no digit is repeated in any such number?

**Solution:**

For a number to be divisible by 5, the last digit should either be 5 or 0.

In this case, 5 is only possible.

For a four digit number to be between 3000 to 4000 , in this case, should start with 3 .

Therefore, the other 2 digits can be arranged by 4 numbers in $P(4,2)$

Formula:

Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is

$P(n, r)=n ! /(n-r) !$

Therefore, a permutation of 4 different objects in 2 places is

$P(4,2)=\frac{4 !}{(4-2) !}$

$=\frac{4 !}{2 !}=\frac{24}{2}=12$

Therefore, there are 12 numbers present between 3000 to 4000 formed by using numbers $3,1,5,6,7,8$.