Wouldn't I use the permutation formula to solve this question: How many positive even integers with at most 3 digits can be created using 2,3,4,5,6, and 7??

Question

anonymous · Answer

no

anonymous · Answer

you have three choices for the last digit

anonymous · Answer

then 5 choices for the tens place and 4 choices for the hundreds place

anonymous · Answer

but that would just be for three digit numbers, it says at least three digits, so there is more work to do

anonymous · Answer

oh no, it says "at most' three digits

anonymous · Answer

for three digit numbers it is $3\times 5\times 4$
for two digit numbers it is $3\times 5$ and for one digit numbers it is 3

anonymous · Answer

add up for your answer

anonymous · Answer

how come only 3 for the hundreds place?

raden · Answer

i think for this case, the repeat numbers is allowed

raden · Answer

so, for 3 digits number
there are 6 * 6 * 3 ways

for 2 digits number
there are  6 * 3 ways

for 1 digit number
there are 3 ways

anonymous · Answer

k I understand. thx again!

raden · Answer

@satellite73 , please verify me :)