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How many three-digit even numbers can be formed by using the digits $0,1,2,3,4,5,6$ if repetition of digits is not permitted?

Initial thoughts: The ones place can only be even so the digits we will be selected will be $\{0,2,4,6\}=$ total of $4$ digits. The second priority is given to the hundreds place as $0$ can't really be selected, so total number of digits gets restricted to $5$ digits. And last priority is given to the ten's place so the total number of choices gets restricted to $6$ digits. So the total number of ways $=6 \cdot 5 \cdot 4=120$. These are my conclusions. Please correct me.

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    $\begingroup$ Note that it makes a difference if the last digit is $0$ or one of $\{2,4,6\}$. I suggest separating that case off. And the part about the tens place doesn't make sense. $\endgroup$
    – lulu
    Commented Sep 1, 2023 at 16:16
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    $\begingroup$ But if the last digit is a $0$, you still have 6 choices for the hundred's digit. And I can't see how you conclude that there are 6 possibilities for the ten's digit. $\endgroup$
    – Henrik supports the community
    Commented Sep 1, 2023 at 16:16

2 Answers 2

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Your reasoning is along the right lines, but there are some small corrections.

First, assume that the ones place has a 0. Then, the hundreds place can have any of {1,2,3,4,5,6}, so there are six choices. Following this, the tens place can have any of the remaining five.

Next, assume the ones place has a 2, 4, or 6. Now, the hundreds place has five choices, since both 0 and one more number are removed. Finally, for the tens place, two options have been removed: the one in the ones place, and the one in the tens place. Thus, there are five remaining options.

So in total, we get $6\cdot 5 + 3\cdot (5\cdot 5) = 105$ options.

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  • $\begingroup$ What is the intuition behind these kind of problems? is there a general abstract idea/concept to it? or is this problem just a concrete one? $\endgroup$
    – 3b1b aimer
    Commented Sep 1, 2023 at 16:30
  • $\begingroup$ I would say that the general idea here is simply counting the number of number of permutations without repetition. The special structure that arises from the fact that 0 is not allowed to be the hundreds digit seems to just be a more concrete example, I am not sure if there is any more general idea behind this $\endgroup$
    – FinleyMarsh
    Commented Sep 1, 2023 at 16:54
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CASE I: if the last digit(units) is even but $0$ may in hundreds digit then the number of arrangements : $6\times 5\times 4=120$

CASE II: if the last digit is even and $0$ is the first(hundreds) digit then the number of arrangements:$1\times 5\times 3=15$

the number of $3$-digit even numbers without repetition is $120-15=105$

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