Permutation and Combination Questions with Answers PDFs for MBA Preparation

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.

Permutation Formula:

ⁿP_r = (n!) / (n-r)!

Combination Formula:

ⁿC_r = n!/r!(

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!

Also Read: MBA Preparation 2025: Tips to Prepare for MBA Entrance Exams

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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Quantitative Aptitude Questions with Answers for Practice

In order to further simplify this concept, we have shared key differences between Permutation and Combination. This table will help you in conceptual clarity.

Read more: Difference between Permutation and Combination

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