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  • ...for students in middle school and high school. The AMC sets the standard in the United States for talented high school students of [[mathematics]]. Th In order of increasing difficulty, AMC competitions are
    5 KB (696 words) - 02:47, 24 December 2019
  • ...E) provides an organized method/formula to find the number of [[element]]s in the [[union]] of a given group of [[set]]s, the size of each set, and the s ...count, in the end making sure every element is counted once and only once. In particular, memorizing a formula for PIE is a bad idea for problem solving.
    9 KB (1,703 words) - 00:20, 7 December 2024
  • In [[combinatorics]], '''constructive counting''' is a [[counting]] technique that involves constructing an item belonging to a set. Along wi ...arity with constructive counting is essential in combinatorics, especially in intermediate competitions.
    13 KB (2,018 words) - 14:31, 10 January 2025
  • ...counting is often a far simpler approach. A large hint that complementary counting may lead to a quick solution is the phrase "not" or "at least" within a pro ...c|</math>, where <math>B^c</math> is the [[complement]] of <math>B</math>. In most instances, though, <math>A</math> is obvious from context and is commi
    8 KB (1,192 words) - 16:20, 16 June 2023
  • ...vertices, each with equal [[probability]]. What is the probability that no two ants arrive at the same vertex? We approach this problem by counting the number of ways ants can do their desired migration, and then multiply this number by the p
    10 KB (1,840 words) - 14:01, 4 July 2024
  • == Solution 1 (Complementary Counting) == ...ath>4</math>-digit integers with at least <math>2</math> or <math>3</math> in it, we can find this by finding the total number of <math>4</math>-digit in
    3 KB (525 words) - 19:25, 30 April 2024
  • ...divisors, it must have <math>4</math> divisors,, so <math>n</math> must be in the form <math>n=p\cdot q</math> or <math>n=p^3</math> for distinct [[prime .../math> and <math>47</math>) so there are <math> {15 \choose 2} =105</math> ways to choose a pair of primes from the list and thus <math>105</math> numbers
    2 KB (249 words) - 08:37, 23 January 2024
  • ...cases: one in which zero is one of the digits and one in which it is not. In the latter case, suppose we pick digits In the second case we choose zero and three other digits such that <math>0<x_2
    3 KB (562 words) - 17:12, 4 March 2022
  • ...th>1</math> and <math> 2^{40} </math> whose binary expansions have exactly two <math>1</math>'s. If a number is chosen at random from <math> S, </math> th ...on is just <math>\binom {40}{2}</math> because we’re choosing 2 1s to go in 40 digit slots. This is equal to 780; we have found <math>q</math>, our den
    8 KB (1,283 words) - 18:19, 8 May 2024
  • .../math> is at least <math>2</math> with a probability that can be expressed in the form <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> a ...B</math> and <math>J</math> differ by <math>1</math> (no zero, because the two numbers are distinct). There are <math>20 \cdot 19 = 380</math> total possi
    5 KB (830 words) - 21:15, 28 December 2023
  • ...een sitting next to each other. If <math>P</math> is written as a fraction in lowest terms, what is the sum of the numerator and denominator? We can use [[complementary counting]], by finding the probability that none of the three knights are sitting ne
    9 KB (1,464 words) - 10:50, 30 October 2024
  • ...<math>4</math>-digit number beginning with <math>1</math> that has exactly two identical digits. How many such numbers are there? Suppose that the two identical digits are both <math>1</math>. Since the thousands digit must be
    5 KB (855 words) - 19:26, 14 January 2023
  • ...Let <math>\frac m n</math> in lowest terms be the [[probability]] that no two birch trees are next to one another. Find <math>m+n</math>. ...oak and maple trees and you also multiply the denominator by the number of ways to arrange the oak and maple trees, making them cancel out.)
    7 KB (1,115 words) - 23:52, 6 September 2023
  • ...octagon meet. How many [[segment]]s joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an [[edge]] or a [[face]]? ...d divide by 2 (because each space diagonal is counted twice because it has two endpoints).
    6 KB (906 words) - 12:25, 19 November 2024
  • ...be tossed <math>10_{}^{}</math> times. Let <math>\frac{i}{j}^{}_{}</math>, in lowest terms, be the [[probability]] that heads never occur on consecutive ...he tails such that no two heads fall between the same tails, and must fall in the positions labeled <math>(H)</math>:
    3 KB (427 words) - 14:03, 24 June 2024
  • ...are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data? ...ly. Also, let <math>t=r+b</math>. The probability <math>P</math> that when two socks are drawn randomly, without replacement, both are red or both are blu
    7 KB (1,328 words) - 19:24, 5 February 2024
  • ...2)</math> in six or fewer steps. Given that <math>p</math> can be written in the form <math>m/n,</math> where <math>m</math> and <math>n</math> are rela ...tt>E</tt>, in some permutation. There are <math>\frac{4!}{2!2!} = 6</math> ways for these four steps of occuring, and the probability is <math>\frac{6}{4^{
    3 KB (602 words) - 22:15, 16 June 2019
  • ...o that each face contains a different number. The [[probability]] that no two consecutive numbers, where <math>8</math> and <math>1</math> are considered ...ertices. The cube has 16 diagonal segments that join nonadjacent vertices. In effect, the problem asks one to count octagonal circuits that can be formed
    11 KB (1,836 words) - 20:08, 26 August 2024
  • ...er get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail ...it strings of <math>0</math>'s and <math>1</math>'s such that there are no two consecutive <math>1</math>'s and no three consecutive <math>0</math>'s.
    15 KB (2,728 words) - 14:36, 23 December 2024
  • ...balanced'' if the sum of its two leftmost [[digit]]s equals the sum of its two rightmost digits. How many balanced integers are there? ...n)^2 = \sum_{n = 1}^9 n^2 = 285</math> balanced numbers. Thus, there are in total <math>330 + 285 = \boxed{615}</math> balanced numbers.
    4 KB (696 words) - 10:55, 10 September 2023

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