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The number of integers greater than 5000 and divisible by 5 that can be formed withthe 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=$$36+120+120=276$$
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