Question

# How many three-digit numbers that are divisible by 5 can be formed using the digits 0, 2, 3, 5, 7, if no digit occurs more than once in each number

A
21
B
20
C
18
D
15

Solution

## The correct option is A 21Since each desired number is divisible by $$5$$, so we must have $$0\ or \ 5$$ at the unit place.Case 1, when $$0$$ is at the units placeThen the hundreds and tens place can be filled up by remaining $$4$$ digits in $$\dfrac { 4! }{ (4-2)! } =\quad \dfrac { 4! }{ 2! } =\quad 4\times 3=\quad 12$$ waysCase 2, when $$5$$ is at the units placeThen the ten's digit can be filled by the digits $$0,2,3,7$$ in $$4$$ waysAnd the $$100's$$ digit can be filled by any of the remaining $$2$$ digits in $$2$$ ways as $$0$$ cannot take Hundred's place, then number wont be a $$3$$-digit number then. So, total number ways $$= 12 + 4 + 2 = 18$$ waysMathematics

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