PnC Formulae



  • Useful Relations - Factorial

    • n! = n.(n-1)!
  • nCr = [nPr] / [r!]

  • Useful Relations - Permutations

    1. nPn = n!

    2. nP0 = 1

    3. nP1 = n

    4. nPn = nPn - 1

    5. nPr = n(n-1Pr-1)

  • Useful Relations - Combinations

    1. nCr = nC(n - r)


      2. Example

      1. 8C6 = 8C2 = {8 x 7}{2 x 1} = 28

    2. nCn = 1

    3. nC0 = 1

    4. nC0 + nC1 + nC2 + ... + nCn = 2n


      5. Example

      1. 4C0 + 4C1 + 4C2 + 4C3+ 4C4 = (1 + 4 + 6 + 4 + 1) = 16 = 24

    5. nCr-1 + nCr =   (n+1)Cr    (Pascal's Law)

    6. If  nCx = nCy  then either x = y or (n-x) = y

  • Selection from identical objects: Some Basic Facts

    1. The number of selections of r objects out of n identical objects is 1
    2. Total number of selections of zero or more objects from n identical objects is n+1.

  • Permutations of Objects when All Objects are Not Distinct

    The number of ways in which n things can be arranged taking them all at a time, when p1 of the things are exactly alike of 1st type, p2 of them are exactly alike of a 2nd type, and pr of them are exactly alike of rth type and the rest of all are distinct is

    {n!}}/{{p1}! x {p2}! ... {pr}!}


  • Permutations with Repetition : Number of permutations of n distinct things taking r at a time (Repetition allowed)

    The number of permutations of n distinct things taking r at a time when each thing may be repeated any number of times is
    nr

  • Circular Permutations: Case 1: when clockwise and anticlockwise arrangements are different

    Number of circular permutations (arrangements) of n different things is
    (n-1)!

  • Circular Permutations: Case 2: when clockwise and anticlockwise arrangements are not different

    Number of circular permutations (arrangements) of n different things, when clockwise and anticlockwise arrangements are not different (i.e., when observations can be made from both sides), is
    [(n-1)!]/[2]

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    S-BATCH