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Permutation and Combination - ( Arithmetic Aptitude)

  1. Factorial Notation: Letnbe a positive integer. Then, factorialn, denotedn! is defined as: n! = n(n - 1)(n - 2) ... 3.2.1. Examples:
    1. We define0! = 1.
    2. 4! = (4 x 3 x 2 x 1) = 24.
    3. 5! = (5 x 4 x 3 x 2 x 1) = 120.
  2. Permutations: The different arrangements of a given number of things by taking some or all at a time, are called permutations. Examples:
    1. All permutations (or arrangements) made with the lettersa,b,cby taking two at a time are (ab,ba,ac,ca,bc,cb).
    2. All permutations made with the lettersa,b,ctaking all at a time are:
      (abc,acb,bac,bca,cab,cba)
  3. Number of Permutations: Number of all permutations ofnthings, takenrat a time, is given by:
    nPr=n(n- 1)(n- 2) ... (n-r+ 1) = n!
    (n-r)!
    Examples:
    1. 6P2= (6 x 5) = 30.
    2. 7P3= (7 x 6 x 5) = 210.
    3. Cor. number of all permutations ofnthings, taken all at a time =n!.
  4. An Important Result: If there arensubjects of whichp1are alike of one kind;p2are alike of another kind;p3are alike of third kind and so on andprare alike ofrthkind,
    such that (p1+p2+ ...pr) =n.
    Then, number of permutations of thesenobjects is = n!
    (p1!).(p2)!.....(pr!)
  5. Combinations: Each of the different groups or selections which can be formed by taking some or all of a number of objects is called acombination. Examples:
    1. Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA. Note: AB and BA represent the same selection.
    2. All the combinations formed bya,b,ctakingab,bc,ca.
    3. The only combination that can be formed of three lettersa,b,ctaken all at a time isabc.
    4. Various groups of 2 out of four persons A, B, C, D are: AB, AC, AD, BC, BD, CD.
    5. Note thatabbaare two different permutations but they represent the same combination.
  6. Number of Combinations: The number of all combinations ofnthings, takenrat a time is:
    nCr= n! = n(n- 1)(n- 2) ... torfactors .
    (r!)(n-r)! r!
    Note:
    1. nCn= 1 andnC0= 1.
    2. nCr=nC(n - r)
    Examples:
    i. 11C4= (11 x 10 x 9 x 8) = 330.
    (4 x 3 x 2 x 1)
    ii. 16C13=16C(16 - 13)=16C3= 16 x 15 x 14 = 16 x 15 x 14 = 560.
    3! 3 x 2 x 1