0
$\begingroup$

Please help! The answer in our textbook is $2$, but I assume that’s a misprint. I answered it myself and I got $12$ possible numbers $(1 \cdot 3 \cdot 1) + (1 \cdot 2 \cdot 2)$ but I just want to make sure :)

“How many $3$ digit even numbers greater than $400$ can be formed from the digits $\{1, 2, 3, 4, 5\}$ if repetition of digits is not allowed?”

$\endgroup$
3
  • $\begingroup$ Hey and Welcome to MSE! Would you care explaining a bit more how did you come up with your solution? $\endgroup$ Feb 25, 2019 at 12:00
  • $\begingroup$ I am also pretty sure that it is a misprint, $2$ seems terribly few $\endgroup$ Feb 25, 2019 at 12:01
  • $\begingroup$ I am sure that your second parenthese is incorrect, how did you get this number? $\endgroup$ Feb 25, 2019 at 12:12

4 Answers 4

1
$\begingroup$

If the number starts with a $4$, it must end with a $2$. Three options for the second digit remain.

If the number starts with a $5$, it can end with a $2$ or a $4$. Again, three options for the second digit remain.

The total number of valid numbers thus equals:

$$3 + 2 \cdot 3 = 9$$

$\endgroup$
1
$\begingroup$

The answer is $9$. The number must start with a $4$ or a $5$ and must end with a $2$ or a $4$, so the possibilities are: $$412, 432, 452, 512, 514, 524, 532, 534, 542.$$

$\endgroup$
0
$\begingroup$

You have two criteria.

  1. The end digit must be $2$ or $4$.
  2. The first digit must be $4$ or $5$.

Just count the ways this works.

$\endgroup$
0
$\begingroup$

You are to create $3$ digit even numbers greater than $400$ so think as follows:

It cannot start with a number less than $4$. So let us start with $4$, then you only have one choice left for the last digit, namely $2$. Can you figure out the rest? (you can peek by mouseovering the yellow boxes, but try to figure out yourself first, before you peek)

$3$ different numbers since the first and last digit is determined and you have $3$ other digits to choose from to fill the middle

It can also start with a $5$, in this case it can end with both $2$ and a $4$. I am sure you can figure the rest out ;)

If it ends with a $2$ you have $3$ choices for the middle number, since once again we fixed the first and the last digit and the same is true for when it ends with a $4$

Adding these together you end up with

a total number of $3+3+3=9$ possibilities.

Hope I could help

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .