Counter
Mathematics Dictionary
Dr. K. G. Shih
Arrange number n as 3 integers


  • Q01 | - Arrange 3 in terms of 3 integers
  • Q02 | - Arrange 4 in terms of 3 integers
  • Q03 | - Arrange 5 in terms of 3 integers
  • Q04 | - Arrange 6 in terms of 3 integers
  • Q05 | - Conclusion

    • Q01. Find arrangement if 3 is expressed in terms of 3 integers

    An integer n and n = n1 + n2 + n2, How many arrangement

    Note
    • Three integers n1, n2, n3 are positive (no zero)
    Projecture a formula for n = 3
    • n1 = 1, n2 = 1 and n3 = 1
    • We have n = 1 + 1 + 1 = 3
    • It has only one arragement.
    • It is the case (n - 2) and has 1 arrangement

    Go to Begin

    Q02. Find arrangement if 4 is expressed in terms of 3 integers

    Note
    • Three integers n1, n2, n3 are positive (no zero)
    • Arrangement : 1, 1, 2 and 2, 1, 1 count as 2 arrangement
    Projecture a formula for n = 4
    • Number 2 stands at 3rd place
      • n = 1 + 1 + 2 = 4
      • It has only one arragement if 2 stands at 3rd place
      • It is the case (n - 2) and has 1 arrangement
    • Number 1 stands at 3rd place
      • 1 + 2 + 1 = 4
      • 2 + 1 + 1 = 4
      • It is the case for number 1 stands at 3rd place
      • Hence it has 2 arragements
      • This is the case (n - 3) and has 2 arrangements
    • Hence there are 3 arrangement
    • It has m terms : m = (n - 2) = 2
    • Sum = m*(m + 1)/2
    • Hence arrangement = 1 + 2 = (n - 1)*(n - 2)/2 = 3

    Go to Begin

    Q03. Find arrangement if 5 is expressed in terms of 3 integers

    Note
    • Three integers n1, n2, n3 are positive (no zero)
    Projecture a formula : n = 5
    • One number is (n - 2) = 3 : it is (n - 2) and only 1 arrangement
      • 1 + 1 + 3 = 5
    • One number is (n - 3) = 2 : it is (n - 3) and only 2 arrangentme
      • 1 + 2 + 2 = 5
      • 2 + 1 + 2 = 5
    • One number is (n - 4) = 1 : it is (n - 4) and only 3 arrangentme
      • 1 + 3 + 1 = 5
      • 3 + 1 + 1 = 5
      • 2 + 2 + 1 = 5
    • Hence arrangement is 1 + 2 + 3 = 6
    Conclusion
    • Number of terms : m = (n - 2) = 3 if n = 5
    • Sum = m*(m + 1)/2 = (n - 2)*(n - 2 + 1)/2 = (n - 1)*(n - 2)/2
    • Hence arranement = (5 - 1)*(5 - 2)/2 = 6
    Note
    • Three integers n1, n2, n3 are positive (no zero)
    Projecture a formula for n = 6
    • One number is (n - 2) = 4 : it is (n - 2) and only 1 arrangement
      • Number 4 stands at 3rd place
      • n = 1 + 1 + 4 = 6. Only one arrangement
    • One number is (n - 3) = 3 : it is (n - 3) and only 2 arrangentme
      • Number 3 stands at 3rd place
      • 1 + 2 + 3 = 6
      • 2 + 1 + 3 = 6
      • It has 2 arrangements
    • One number is (n - 4) = 2 : it is (n - 4) and only 3 arrangentme
      • Number 3 stands at 3rd place
      • 1 + 3 + 2 = 6
      • 3 + 1 + 2 = 6
      • 2 + 2 + 2 = 6
      • It has 3 arrangements
    • One number is (n - 5) = 1 : it is (n - 5) and only 3 arrangentme
      • 1 + 4 + 1 = 6
      • 4 + 1 + 1 = 6
      • 2 + 3 + 1 = 6
      • 3 + 2 + 1 = 6
      • It has 4 arrangenments
    Conclusion
    • Hence arranements = 1 + 2 + 3 + 4 = (6 - 1)*(6 - 2)/2 = 10
    • Last term = T(m) = (n - 2) = 4
    • Number of terms m = (n - 2) = 4
    • Sum = m*(m + 1)/2 = (n - 1)*(n - 2)/2

    Go to Begin

    Q04. Find arrangement if 6 is expressed in terms of 3 integers

    Note
    • Three integers n1, n2, n3 are positive (no zero)
    Projecture a formula for n = 6
    • One number is (n - 2) = 4 : it is (n - 2) and only 1 arrangement
      • Number 4 stands at 3rd place
      • n = 1 + 1 + 4 = 6. Only one arrangement
    • One number is (n - 3) = 3 : it is (n - 3) and only 2 arrangentme
      • Number 3 stands at 3rd place
      • 1 + 2 + 3 = 6
      • 2 + 1 + 3 = 6
      • It has 2 arrangements
    • One number is (n - 4) = 2 : it is (n - 4) and only 3 arrangentme
      • Number 2 stands at 3rd place
      • 1 + 3 + 2 = 6
      • 3 + 1 + 2 = 6
      • 2 + 2 + 2 = 6
      • It has 3 arrangements
    • One number is (n - 5) = 1 : it is (n - 5) and only 3 arrangentme
      • Number 1 stands at 3rd place
      • 1 + 4 + 1 = 6
      • 4 + 1 + 1 = 6
      • 2 + 3 + 1 = 6
      • 3 + 2 + 1 = 6
      • It has 4 arrangenments
    Conclusion
    • Hence arranements = 1 + 2 + 3 + 4 = (6 - 1)*(6 - 2)/2 = 10
    • Number of terms m = (n - 2) = 4
    • Sum = m*(m + 1)/2 = (n - 1)*(n - 2)/2

    Go to Begin

    Q05. Conclusion

    Conclusion
    • For (n - 2), it has 1 arrangement
    • For (n - 3), it has 2 arrangements
    • For (n - 4), it has 3 arrangements
    • Etc.
    • Number of terms : m = (n - 2)
    • Sum = m*(m + 1)/2
    • Hence total arrangements = 1 + 2 + 3 + ..... + (n - 2) = (n - 2)*(n - 1)/2
    • The formula is C(n-1, 2) = (n - 1)*(n - 2)/(2!)

    Go to Begin
    Show Room of MD2002 Contact Dr. Shih Math Examples Room
    Copyright © Dr. K. G. Shih, Nova Scotia, Canada.
    1