My question reads as follows

Find the number of passwords that use each of the digits 3,4,5,6,7,8,93, 4, 5, 6, 7, 8, 9 exactly once

In how many of the passwords

(a) are the first three digits even?

(b) are the three even digits consecutive?

(c) are the four odd digits consecutive?

(d) are no two odd digits consecutive?

Now for the first part I got 7!7! And then moved on to a. I did 3!4!3!4! and got 144144 but I am not too sure this is correct.

For b through d I’m confused as to what is meant by consecutive.

When I attempted b I got something like 7!−5(4!)7!-5(4!) because I took the total subtracted my five cases where we can have 468468 and the multiplied by 4!4! Since the odds still need to be arranged. I am not sure if my technique is correct so this is confusing me for c and d.

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For the first, count possible permutations between 3,4,5,6,7,83,4,5,6,7,8 and99.

– Abdallah Hammam

Oct 20 at 22:18

Consecutive means that they appear right after one another. One example for bb would be 46835794683579 and another would be 35684973568497. However, 34659873465987 wouldn’t work because 88 is not adjacent to 44 or 66.

– Kevin Long

Oct 20 at 22:26

@KevinLong so I can have say 486 not just 468 in that order?

– Sam

Oct 20 at 22:27

@Correct, you can have any permutation of 468468 in your password, as long as there are no odd numbers in between the evens.

– Kevin Long

Oct 20 at 22:27

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1 Answer

1

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Your answers for the number of passwords and part a are correct. Now you need to allow your three consecutive even numbers to take any place in the password, not just the first three places, as long as they remain consecutive. So you still have 3!∗4!3!*4!, but now you take into account the number of places the even numbers can take. What you have right now (7!−5∗4!)7!-5*4!) counts the number of passwords that don’t have 468468 occurring in any place. Part c is similar to part b, but with the odd numbers. For part d, note that you have more odd numbers than even numbers (exactly one more, to be precise). How can you arrange them to get a password where no two odd numbers are adjacent?

I worked out b and got 5(3!)(4!)=720. C is similar but not its four digits since there are four odd so I have 4(4!)(3!)=576. For d I worked out a chart and got 15 possible combinations thus 15(4!)(3!). I’m not too sure if theses are fine.

– Sam

Oct 20 at 22:49

You’re right for b- you have 3!3! way to order the evens, 4!4! ways to order the odds, and 55 ways to arrange the odds around the evens (00 to the left and 55 to the right, 11 to the left and 44 to the right, up to 55 to the left and 00 to the right). You’re also right for c, since you can have 0,1,20,1,2 or 33 of your evens to the left of your odds, and the rest to the right. I’m not sure where you got that 1515 in part dd, however. If my first number is odd, the next must be even, then odd, etc, so we get oeoeoeooeoeoeo, where oo is odd, ee is even. Can my password start with an odd in d?

– Kevin Long

Oct 20 at 22:54

I did odd even odd then odd even even odd, then odd even even even odd, and so on and then switched to odd in the second place and so on. This is how I got 15.

– Sam

Oct 20 at 23:01

Oh. Wait. I just considered two odds not the other two that would also have to be non consecutive.

– Sam

Oct 20 at 23:09