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?


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) =


= = = 12.


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


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