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How to find the number of distinct four digit numbers that are increasing or decreasing?

The correct answer is $2{9 \choose 4} + {9 \choose 3} = 343$. How to get there?

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For every collection of $4$ distinct digits, there is a unique way to arrange the digits so that they are decreasing, and a unique way to arrange them so that they are increasing. This gives $2 \binom{10}{4}$ different sequences that are increasing or decreasing, but not all correspond to a four-digit number. For increasing sequences, you cannot have $0$ as a first digit. This rules out $\binom{9}{3} = 84$ possibilities. So the total is $2(210) - 84=336 \neq 343$...?

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The analysis that was used goes as follows:

Not using $0$: We choose $4$ non-zero digits. Once we have done that, we can arrange them in increasing order in $1$ way, and in decreasing order in $1$ way, for a total of $2\binom{9}{4}$.

Using $0$: They can only be decreasing. And we need to choose $3$ non-zero digits to go with the $0$. This can be done in $\binom{9}{3}$ ways.

I prefer Juanito's approach. Note that the sum is not $343$, so if the book got that, there is a computational error.

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Strange, my textbook shows 343 and why is n = 9? Is it just 1 -9 or is there more meaning behind using 9 as n. I am probably overthinking this. – user3718584 Jul 21 '14 at 17:56
There are $9$ non-zero digits, so yes, it is $1,2,\dots,9$. – André Nicolas Jul 21 '14 at 17:58
Why is it that we are not allowed to start with 0 for an increasing 4 digit set? – user3718584 Jul 21 '14 at 18:01
Because we are using the $9000$ four-digit numbers. The number $0347$ is not a four-digit number. – André Nicolas Jul 21 '14 at 18:04

For decreasing numbers, we can also take zero as a possible candidate digit( $4320$ is ok, $0234$ not ok). Once we have a set of 4 digits(for example, $1-9-6-7$), the number is uniquely determined by the increasing/ decreasing order($1679$ or $9761$).

So, number of decreasing numbers

= numbers without $0$ as a digit + numbers with $0$ as one of the digits.

= all possible combinations of 4 digits from $\{1, ... 9 \}$ + all possible combinations of $3$ digits from $\{1, ... 9\}$, $0$ being the already selected digit.

= ${9 \choose 4} + {9 \choose 3}$.

Number of increasing numbers=all possible combinations of 4 digits from $\{1, ...9\} = {9 \choose 4}$.

Add those two.

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I'm not entirely following the decreasing part, im having trouble interpreting the sentence sorry. – user3718584 Jul 21 '14 at 17:52
Is it clearer now? Let me know, accordingly I can try to make some more edits. – Juanito Jul 21 '14 at 17:56
Thank you, more clear now. – user3718584 Jul 21 '14 at 18:01

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