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AMC 8 2020 Problem 7 | Counting Problem

Try this beautiful Problem based on combinatorics from AMC 8 2020.

Counting Problem - AMC 8 2020 Problem 7


How many integers between 2020 and 2400 have four distinct digits arranged in increasing order? (For example, 2347 is one integer.

  • 9
  • 10
  • 15
  • 21
  • 28

Key Concepts


Combinatorics

Counting

Suggested Book | Source | Answer


AMC 8 2020 Problem 7

15

Try with Hints


The second digit can't be 1 or 2, since the digit need to be increasing and distinct , and the second digit can't be 4 also since the number need to be less than 2400, so its 3

now we need to choose the last two digit from the set \{4,5,6,7,8,9\}

now we can do it in 6C2= 15 ways. now in only one way we can order so there are 15 numbers.

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Try this beautiful Problem based on combinatorics from AMC 8 2020.

Counting Problem - AMC 8 2020 Problem 7


How many integers between 2020 and 2400 have four distinct digits arranged in increasing order? (For example, 2347 is one integer.

  • 9
  • 10
  • 15
  • 21
  • 28

Key Concepts


Combinatorics

Counting

Suggested Book | Source | Answer


AMC 8 2020 Problem 7

15

Try with Hints


The second digit can't be 1 or 2, since the digit need to be increasing and distinct , and the second digit can't be 4 also since the number need to be less than 2400, so its 3

now we need to choose the last two digit from the set \{4,5,6,7,8,9\}

now we can do it in 6C2= 15 ways. now in only one way we can order so there are 15 numbers.

Subscribe to Cheenta at Youtube


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