All the Quant, page 410: Office manager combinatorics

[yo4561]

For the following question: "An office manager must choose a four-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of digits is allowed. How many different lock codes are possible?"

The answer is: 5*8*7*4.

I am confused as to how you guarantee there are no repetitions. I realize that you reduce the middle slots to reduce the repetition, but what is preventing the numbers from all being let's say 3 even when you reduce the numbers? Also, why do you start with the first slot and then the last slot... is it because those are more restrictive? Initially, I did 5*9*8*2 thinking that the second number could be any number but the one chosen for the first slot, the third spot cannot be the first two, and for the fourth spot... I have to account for the two previous spots.

I am very confused. Any insights would be so appreciated.

[esledge]

Also, why do you start with the first slot and then the last slot... is it because those are more restrictive?

Yep, I think that's exactly it.

If you go from first to last, things are easy when you pick the first digit, which has to be odd: 5 possibilities.
You are ok when you pick the second digit, which just can't be the same as the first: 9 possibilities.
Same for the third digit (just can't be the same as the first two): 8 possibilities.

But then when you select the last digit, you have a bunch of unknown scenarios that must be considered separately:
IF all of the first three digits are odd, then there's only 2 possibilities left for the last digit.
IF two of the first three digits are odd, then there's only 3 possibilities left for the last digit.
If just the first of the first three digits was odd, then there's 4 possibilities for the last digit.

So this would require you to count all of those scenarios separately and start thinking of this as an AND and an OR problem, but it's easier to leave it as an AND problem (pick the first odd AND then the last odd AND then the middle two remaining numbers) with the more restrictive selections made first.