# 2006 Alabama ARML TST Problems/Problem 4

## Problem

Find the number of six-digit positive integers for which the digits are in increasing order.

NOTE: Increasing from left to right. Digits are distinct.

## Solution

### Solution 1

Think of a nine-digit number $123456789$. If you take out $3$ digits, then it will become a $6$-digit number and all the digits will still be in increasing order. The number of ways to take three digits out is $\binom{9}{3}=\boxed{84}$

### Solution 2

We pick six different digits $a$ through $f$ for the integer. None of them can be 0, or else it is a five digit integer or the digits are not in increasing order. Let's say that $a$ is the least digit of them all. $a$ is therefore the hundred-thousands digit. Let's say that $b$ is the second smallest integer. Then $b$ is the ten-thousands digit. etc.

For each group of $a-f$ we pick, there is only one arrangement such that each digit is in increasing order. There are $\binom{9}{6}=\binom{9}{9-6=3}=\boxed{84}$ ways to pick the digits, therefore there are 84 integers.

## See also

 2006 Alabama ARML TST (Problems) Preceded by:Problem 3 Followed by:Problem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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