Difference between revisions of "1992 AIME Problems/Problem 2"

Latest revision as of 04:18, 4 November 2022

Problem

A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there?

Solution 1

Note that an ascending number is exactly determined by its digits: for any set of digits (not including 0, since the only position for 0 is at the leftmost end of the number, i.e. a leading 0), there is exactly one ascending number with those digits.

So, there are nine digits that may be used: $1,2,3,4,5,6,7,8,9.$ Note that each digit may be present or may not be present. Hence, there are $2^9=512$ potential ascending numbers, one for each subset of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ .

However, we've counted one-digit numbers and the empty set, so we must subtract them off to get our answer, $512-10=\boxed{502}.$

Video Solution by OmegaLearn

https://youtu.be/vxTBkiHKfdY?t=31

~ pi_is_3.14

Solution 2

We see that we can express the number of ways for a 2 digit number as $\binom{8}{1}+\binom{7}{1}...\binom{1}{1}$ and a 3 digit number as $\binom{8}{2}+\binom{7}{2}+...\binom{2}{2}$ this is because once we choose the first number the next two numbers can be any two of the numbers above it. Using the hockey stick identity repeatedly we can get $\binom{9}{2}+\binom{9}{3}+...+\binom{9}{9}$ . We know that $\binom{n}{0}+\binom{n}{1}+...\binom{n}{n}=2^n$ this can be proved easily by story. If you have n people and you want to know how many ways you can make a group of any size you could do this $2$ ways. You could count everyone is binary in or out so $2^n$ or you could count each case individually: $\binom{n}{0}+\binom{n}{1}+...\binom{n}{n}$ so the 2 statements are equivalent. Therfore we have $2^9-\binom{9}{1}-\binom{9}{0}=\boxed{502}$ Note how you could have skipped the hockey stick step by choosing all the numbers first.

1992 AIME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 8: / Line 8: @@
 However, we've counted one-digit numbers and the [[empty set]], so we must subtract them off to get our answer, <math>512-10=\boxed{502}.</math>
+== Video Solution by OmegaLearn ==
+https://youtu.be/vxTBkiHKfdY?t=31
+~ pi_is_3.14
 == Solution 2 ==
-We see that we can express the number of ways for a 2 digit number as <math>\binom{8}{1}+\binom{7}{1}...\binom{1}{1}</math> and a 3 digit number as <math>\binom{8}{2}+\binom{7}{2}+...\binom{2}{2}</math> this is because once we choose the first number the next two numbers can be any two of the numbers above it. Using the hockey stick identity repeatedly we can get <math>\binom{9}{2}+\binom{9}{3}+...+\binom{9}{9}</math>. We know that <math>\binom{n}{0}+\binom{n}{1}+...\binom{n}{n}=2^n</math> this can be proved easily by story. [hide=proof]If you have n people and you want to know how many ways you can make a group of any size you could do this <math>2</math> ways. You could count everyone is binary in or out so <math>2^n</math> or you could count each case individually: <math>\binom{n}{0}+\binom{n}{1}+...\binom{n}{n}</math> so the 2 statements are equivalent.[/hide]therfore we have <math>2^9-(\binom{9}{1}+\binom{9}{0})=\boxed{502}</math>
+We see that we can express the number of ways for a 2 digit number as <math>\binom{8}{1}+\binom{7}{1}...\binom{1}{1}</math> and a 3 digit number as <math>\binom{8}{2}+\binom{7}{2}+...\binom{2}{2}</math> this is because once we choose the first number the next two numbers can be any two of the numbers above it. Using the hockey stick identity repeatedly we can get <math>\binom{9}{2}+\binom{9}{3}+...+\binom{9}{9}</math>. We know that <math>\binom{n}{0}+\binom{n}{1}+...\binom{n}{n}=2^n</math> this can be proved easily by story. If you have n people and you want to know how many ways you can make a group of any size you could do this <math>2</math> ways. You could count everyone is binary in or out so <math>2^n</math> or you could count each case individually: <math>\binom{n}{0}+\binom{n}{1}+...\binom{n}{n}</math> so the 2 statements are equivalent. Therfore we have <math>2^9-\binom{9}{1}-\binom{9}{0}=\boxed{502}</math>
 Note how you could have skipped the hockey stick step by choosing all the numbers first.
 {{AIME box|year=1992|num-b=1|num-a=3}}
 {{MAA Notice}}

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