2006 Alabama ARML TST Problems/Problem 4


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 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
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