The number of integers greater than 5000 and divisible by 5 that can be formed with the digits 1, 3, 5, 7, 8, 9 where no digit is repeated is

We are being given with six digits- 1,3,5,7,8,9

So for a number formed by these digits to be greater than 5000, the number can be either 4-digit,5-digit or 6-digit

i.)4-digit number:

Since the number is divisible by 5, so last digit has to be 5

Also, the number is greater than 5000 so first digit has to be 7,8 or 9

So, the first digit can be selected in $$^3C_1$$ ways

For the remaining 2-places, we can select 2 digits from the remaining 4 digits in $$^4C_2$$ ways and arrange them in 2! ways

So, number of ways of forming the 4-digit number=$$^3C_1\times^4C_2\times\ 2!=36$$ ways

ii.)5-digit number:

The last digit has to be 5 as the number is divisible by 5

Now, for the remaining 4 places we have to select 4 digits from 5 digits which can be done in $$^5C_4$$ ways and rearrange them in $$4!$$ ways

So, number of ways of forming the 5-digit number=$$^5C_4\times\ 4!=120$$ ways

iii.)6-digit number:
The last digit has to be 5 as the number is divisible by 5

So, the remaining 5 places can be filled by remaining 5 digits in $$5!=120$$ ways

So, number of ways of forming the 6-digit number=$$5!=120$$ ways

So, number of integers=$$36+120+120=276$$