Difference between revisions of "2006 Alabama ARML TST Problems/Problem 4"

Revision as of 01:32, 12 December 2011

Problem

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

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.

@@ Line 3: / Line 3: @@
 ==Solution==
+===Solution 1===
+Think of a nine-digit number <math>123456789</math>. If you take out <math>3</math> digits, then it will become a <math>6</math>-digit number and all the digits will still be in increasing order. The number of ways to take three digits out is <math>\binom{9}{3}=\boxed{84}</math>
+===Solution 2===
 We pick six different digits <math>a</math> through <math>f</math> 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 <math>a</math> is the least digit of them all. <math>a</math> is therefore the hundred-thousands digit. Let's say that <math>b</math> is the second smallest integer. Then <math>b</math> is the ten-thousands digit. etc.

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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