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How many $3$ digit different number that will be divisible by $5$ can be formed from the digit $0,2,3,4,5,6$ lying between $100$ and $1000$.


My attempt:

Divisible by $5$ is possible only when unit digit will be either $0$ or $5$, means possibility for unit digit is $2$.

And possible number of choices for third place digit is $5$ (except $0$).

And possible number of choices for tenth place digit is $6$ (all).

So, total possible such numbers are $=5\times6\times2=60$ (answer).

Can you explain in formal/alternative way, please?

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Here is a formal way (though not much different than yours)...


Add up the following:

  • The amount of numbers of the form $\underbrace{2,3,4,5,6}_{5\text{ options}}|\underbrace{0,2,3,4,5,6}_{6\text{ options}}|\underbrace{0}_{1\text{ option}}$
  • The amount of numbers of the form $\underbrace{2,3,4,5,6}_{5\text{ options}}|\underbrace{0,2,3,4,5,6}_{6\text{ options}}|\underbrace{5}_{1\text{ option}}$

The total amount is $5\cdot6\cdot1+5\cdot6\cdot1=60$.

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  • $\begingroup$ Thanks for nice explanation. $\endgroup$ – 1 0 May 29 '16 at 7:19
  • $\begingroup$ @MithleshUpadhyay: You're welcome :) $\endgroup$ – barak manos May 29 '16 at 7:22

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