Tricks to Solve Permutation and Combination Questions

Permutations and combinations are essential topics in probability and quantitative aptitude, widely applicable in both competitive exams and everyday problem-solving. Mastering these concepts will help you solve problems related to arrangements and selections, which are common in various competitive exams like SBI PO, SBI CLERK, IBPS PO, IBPS CLERK, RRB PO, NICL AO, LIC AAO, SNAP, MAT, SSC CGL, college placements and more.

In this article, we will cover various tips, tricks, and shortcuts to solve permutations and combinations questions.

Tips and Tricks for Permutation and Combination

• The factorial of 1 is always 1 (1! = 1).
• The factorial of 0 is always 1 (0! = 1).

• The solution to a permutation problem involves the following two steps:
  • Step 1: Identifying the objects and the positions involved in the problem.
  • Step 2: Determining whether repetition of objects is permitted or not.

• In permutations, we only need the first r terms of n!, stopping at (n − r + 1).
  • Example: The number of ways to arrange r items from a set of n = 7:
    • P(7, 1) = 7 = 7
    • P(7, 2) = 7 × 6 = 42
    • P(7, 3) = 7 × 6 × 5 = 210
    • P(7, 4) = 7 × 6 × 5 × 4 = 840
    • P(7, 5) = 7 × 6 × 5 × 4 × 3 = 2520
    • P(7, 6) = 7 × 6 × 5 × 4 × 3 × 2 = 5040
    • P(7, 7) = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

• Simplified Formula for Combinations: Only consider the first r terms in the product.
  • For combinations, use the formula: C(n, r) = n × (n − 1) × (n − 2) … (n − r + 1)/ r!
    • C(8, 1) = 8/1 ​= 8
    • C(8, 2) = (8 × 7)/( 2 × 1) ​= 56/2 ​= 28
    • C(8, 3) = (8 × 7 × 6)/( 3 × 2 × 1) ​= 336/6 ​= 56
    • C(8, 4) = (8 × 7 × 6 × 5 )/(4 × 3 × 2 × 1) = 1680/24 = 70
    • C(8, 5) = (8 × 7 × 6 × 5 × 4)/(5 × 4 × 3 × 2 × 1) = 6720/120 ​= 56
    • C(8, 6) = (8 × 7 × 6 × 5 × 4 × 3)/(6 × 5 × 4 × 3 × 2 × 1) = 20160/720 =28
    • C(8, 7) = (8 × 7 × 6 × 5 × 4 × 3 × 2)/(7 × 6 × 5 × 4 × 3 × 2 × 1) = 40320/5040 =8
    • C(8, 8) = 1

• When you see clue words like arrangement, schedule, or order, it usually indicates that the problem involves permutations.

• When the order is important use permutation.

• Every permutation involves order, but not every combination involves order.

• The relationship between permutation and combination can be expressed as P(n, r) = C(n, r) × r!

• The number of ways to arrange r objects from n distinct objects, where repetition is allowed, is n^r.

• If some objects are identical, the formula for permutations changes: n! / k₁!k₂!k₃!...
  • Where k₁, k₂or,… are the frequencies of identical objects.

• When m specified objects must always come together in a permutation, treat those m objects as a single "block" or "unit," and the number of permutations of n objects is given by m! × (n − m + 1)!

• When you see clue words like group, committee, sample, selection, or subset, it usually indicates that the problem involves combinations.

• When the order is not important use combination.

• Use the symmetry property of combinations: C(n, r) = C(n, n − r).
  • Example: C(10, 3) = C(10,7), so you can choose the smaller value to reduce computation.

• If r = 1 or r = n, then C(n, 1) = n and C(n, n) = 1.
  • Example: C(7, 1) = 7 and C(7, 7) = 1.

• Combinations with Repetition: For selecting r objects from n types with repetition allowed, the formula is:
  • C(n + r − 1, r)

• When n and r are the same, the number of permutations will always be greater than the number of combinations.
• As the order does not matter in combinations, selecting the same k objects from a set of n elements will always result in the same outcome.

Articles Related to Permutations and Combinations:

• Permutaions and Combinations.
• Apptitude Questions on Permutations and Combinations.
• Quiz about Permutations and Combinations.