Permutation and Combination: Theorems, Sample Questions, Tips and FAQs

The counting rule or the fundamental counting principle states that if an event has p outcomes and another event has q outcomes, then the number of outcomes when both these events occur together is p x q.

A Permutation is an arrangement in a particular order of several things considered a few or all at a time.

Theorem 1: The number of permutations of k different objects taken l at a time, where 0 ^kP_l

Theorem 2: The number of permutations of k different objects taken l at a time, where repetition is allowed, is k^r.

Theorem 3: The number of permutations of k objects, where l objects are of the same kind and the rest is all different is k!/l!

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A Combination refers to the number of possible arrangements possible for an event. In combination, the order of selection does not matter.

The number of combinations of k items taken at a time is denoted as ^kC_l.

Theorem 4: ^kC_k-l = k!/(k-l)!(k- (k-l))! = k!/(k-l)! l!= ^kC_l

We infer that, selecting l items from k items also means rejecting k-l items.

Theorem 5: ^kC_l = ^kC^m ⇒ l = m or l = k-m, i.e., k = l + m

Theorem 6: ^kC_l = ^kC_l-1 = ^k+1C_l

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Illustrated Examples of Permuation-Combination

Q 1: Find the value of k such that, ^kP₅ = 42 ^kP₃ , k>4

Solution:

Given that, ^kP₅ = 42 ^kP₃

Applying the formula of permutation,

k (k – 1) (k – 2) (k – 3) (k– 4) = 42 k(k – 1) (k– 2) 

According to the question, k > 4 so k(k – 1) (k – 2) ≠ 0

Therefore, by dividing both sides by k(k – 1) (k – 2), we get

(k – 3 (k – 4) = 42 or k2 – 7k – 30 = 0 

or k2 – 10k + 3k – 30 or (k – 10) (k + 3) = 0 

or k – 10 = 0 or k + 3 = 0 or k = 10  or k = – 3

As k cannot be negative, so k= 10.

Q 2: If ^kC₉ = ^kC₈, find ^kC₁₇ ?

Solution:

Given, ^kC₉ = ^kC₈

k!/9!(k-9)!= k!/8! (k-8)!

That is, 1/9= 1/k-8

So, k-8 = 9

Hence, k=17 and ^kC₁₇ = ¹⁷C₁₇ = 1

Q 3: In how many ways can 5 balls and 3 boxes be arranged in a row so that no two boxes are together?

Solution:

The 5 balls can be arranged in 5! ways, and to ensure that no two boxes are together, 3 boxes can be put in ⁶C₃ ways.

Total = 5! X  ⁶C₃ = 5! X 6!/3!= 4x 5x 2x 3x 4x 5x 6 = 14,400 ways

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