#1**+5 **

An even three digit number will have a "2" or a '6" at the end.

So, if it has a "2" at the end, we just want to find the number of permutes that can be formed by selecting 2 of any of the remaining 4 digits= nPr(4,2) = 12

And the same number can be made if the "6" is at the end, so that's 12 + 12 = 24

The number of even four digit numbers is similar. The number will have a 2 or 6 at the end, and we just want to find the number of permutes that can be formed by selecting any 3 of the remaining 4 digits = nPr(4,3) = 24. And, as above, the same number can be formed if the 6 is at the end.

So, since "or" usually means "add," we have 12 + 12 + 24 + 24 = 72 total even 3 or 4 digit even numbers

CPhill Jan 28, 2015

#1**+5 **

Best Answer

An even three digit number will have a "2" or a '6" at the end.

So, if it has a "2" at the end, we just want to find the number of permutes that can be formed by selecting 2 of any of the remaining 4 digits= nPr(4,2) = 12

And the same number can be made if the "6" is at the end, so that's 12 + 12 = 24

The number of even four digit numbers is similar. The number will have a 2 or 6 at the end, and we just want to find the number of permutes that can be formed by selecting any 3 of the remaining 4 digits = nPr(4,3) = 24. And, as above, the same number can be formed if the 6 is at the end.

So, since "or" usually means "add," we have 12 + 12 + 24 + 24 = 72 total even 3 or 4 digit even numbers

CPhill Jan 28, 2015