Ytukyg

Number of ways to form 3 digit even numbers from 1,2,5,6,9 if each digit can only be used once.

my understanding is that for even numbers, the last digit must be even or 0.

picking 3 numbers ->

my last pick (3rd number) i understand that i only need to have 2 available cards to pick and these 2 cards needs to be an even number (2,6)

so working backwards, my first pick will have 4 numbers to choose from,

So $4 cdot 3 cdot 2 $ is the answer.

But having 5 numbers to choose from, why is the first pick only 4 different possible ways ?

But having 5 numbers to choose from, why is the first pick only 4 different possible ways ?

Because there aren't five possible numbers to choose from. One of them you have already used in the last position.

You started out correctly by first focusing on the last digit, for which there are indeed $3$ options. So, after you have picked one of those, there are $4$ digits left to pick from for the first digit, and then there are $3$ choices left for the second digit. So: $3cdot 4 cdot3$ possibilities

So, the correct answer is not $4cdot 3 cdot 2$

Moreover, the choice for the first digit is not $5$ if you first pick one of the $3$ even numbers for the last digit first.

Now, you could start by picking one of the $5$ Numbers for the first digit first, but the issue is that if you do that, you eill need to distinguish between the case where you picked an even number for that first digit or an odd number, and now you get a more complicated equation

Pick the last digit first. You have $3$ choices ($2,6$ or $8$).

Then pick the first digit. As you have already picked a digit you have only $4$ choices remaining.

Then pick the second digit. As you have already picked two you have only three choices remaining.

So the number of ways to do this is $3*4*3 = 36$

If you don't pick the last digit first you will have $5$ choices for the first choice and $4$ for the second but when coming to choosing the last digit you do not know if you have $3$ choices, $2$ chooses or only $1$. So you can't do it that way without tricky subcases (which can be done but is trickier). ANd it doesn't matter which order you choose the digits. So it's a good idea to pick the most restrictive digit first.

Maybe the following make it clearer.

$overbrace{begin{cases}overbrace{begin{cases} \
overbrace{begin{cases} 152\162\182
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 512\562\582
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 612\652\682
end{cases}}^{text{first digit is 6 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 812\852\862
end{cases}}^{text{first digit is 8 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 2 (there are 4 choices for the first digit)}}\
overbrace{begin{cases} \
overbrace{begin{cases} 156\126\186
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 216\256\286
end{cases}}^{text{first digit is 2 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 516\526\586
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 816\826\856
end{cases}}^{text{first digit is 8 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 6 (there are 4 choices for the first digit)}}\
overbrace{begin{cases} \
overbrace{begin{cases} 128\158\168
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 218\258\268
end{cases}}^{text{first digit is 2 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 518\528\568
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 618\628\658
end{cases}}^{text{first digit is 6 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 8 (there are 4 choices for the first digit)}}end{cases}}^{text{There are 3 choices for the last digit; either 2,6, or 8}}$