Question

A 5 digit passcode is to be used for entry into an office building. Francis plans to select her passcode in such a way that the 5 digit code ends with an even digit, has all 5 digits distinct, and does not contain any digit greater than 6. How many ways can Francis select her code?

Answer 1

Ways that last digit can be even = 4, since 0,2,4 & 6 are the only possibilities
There are 6 numbers total available after the last digit is chosen.
The number of distinct 4-digit sequences available out of the 6 available is
6*5*4*3

So, total number of ways is 6*5*4*3 * 4 = 1440

Answer 2

there are 3 even numbers < 6: 0, 2, 4
there are 6 numbers < 6: 0, 1, 2, 3, 4, 5

To solve work from the last digit forward
For the first choice (last pass number in code), she has 3 choices: 0, 2, 4
(that leaves 5 total digits remaining: 2 evens digits & 1, 3, 5)
For the second pass number, she has 5 choices: 6-first number chosen
For the third pass number, she has 4 choices: 6-first two numbers
etc
.
.
.
so the total number of pass key combinations is 3*5*4*3*2=360 possibilities

Answer 3

I believe that 6 is a possible outcome.

Answer 4

yes i did not read carefully, le 6

Answer 5

so the product would be 4*7*6*5*4=3360
4 evens
7 digits total

@ybarrap, do you agree with this answer?

Answer 6

Yes, there are 7 digits total. However, after selecting the even at the end, we have 6 numbers available, then after selecting this one we have 5 and so on. so number of was is 4 * 6*5*4*3.

Answer 7

i quit on this one! logic good,execution baaaaaad :{}