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Can someone take a look at my answers to these questions? I'm primarily concerned with my second answer. I believe these answers to be correct but am wondering if this is the most elegant approach or if there may be another approach one would use.

This problem concerns lists made from the symbols $A,B,C,D,E,F,G,H,I$.

(a) How many length-$5$ lists can be made if repetition is not allowed and the list is in alphabetical order?

My answer:

$$9 \choose 5$$

Since this counts the number of unique $5$-element subsets we can form from a $9$ element set and each subset has one order.

(b) How many length-$5$ lists can be made if repetition is not allowed and the list is not in alphabetical order?

My answer:

$${9 \choose 5 }\cdot 5! - {9 \choose 5}$$

My thought here was we can select our $5$-element subsets from a $9$ element set in $9 \choose 5$ ways and permute those elements in $5!$ ways. But we must remove each ordering in which they are alphabetized (which from part (a) we know is $9 \choose 5$).

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  • $\begingroup$ When you say "list" then the reader thinks of order; "the first element of the list", "the second element of the list", etc. Your answer hints that order is not important. $\endgroup$
    – zoli
    Commented Dec 5, 2016 at 22:34
  • $\begingroup$ @zoli I've put my justifications below the answers. For the first answer if you are selecting 5 elements from a 9 element subset, each subset is unique and has exactly one order that is alphabetized. The second answer selects 5 elements and permutes them (counting all of the other orders) and removes the alphabetized orders previously counted. $\endgroup$
    – ClownInTheMoon
    Commented Dec 5, 2016 at 22:37
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    $\begingroup$ Perfect. And the shortest, simplest way. And your justifications are clear and concise. $\endgroup$
    – DanielWainfleet
    Commented Dec 5, 2016 at 22:39

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Your answer is correct in formal language that 'not' means 'explicitly not'.

Though often in the natural language phrasing of this question, 'not' means 'not necessarily'.

For clarity, I would have expected this in the question already, but since that is not the case, I would state this in the answer.

Also ${9 \choose 5} (5! -1)$ would reflect this better: any selection of 5, times any ordering except the alphabetical one.

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