Problems on Permutations and Combinations : CAT Quantitative Aptitude

Let’s look at some application based problems in Permutations and Combinations. We will take you through each step of the solution but while writing the exam it is advisable to just get the answer with minimal calculation steps.

Example:

A man has 10 friends whom he wants to invite for dinner. The number of ways in which he can invite at least one of them is:

[1] 1024

[2] 10!

[3] 1023

[4] 10! - 1

Solution:
Suppose the man has to invite only one of his friends (out of 10) for dinner, then we know that he has to choose one amongst his 10 friends. This can be done in: 10_C1 ways

The key word in the question is: at least, this means the friend can call 1 or 2 or 3 or…10 friends over for dinner. Mathematically, this can be shown as:

 
 
Therefore, the man can invite his 10 friends in (1024 – 1) or 1023 ways.

Example:
Assume the above problem is framed as follows:

A man has 10 friends whom he wants to invite for dinner. The number of ways in which he can invite at least three of them is:

(1) 1021                         

(2) 968

(3) 864                          

(4) None of these

Solution:

Suppose the man has to invite only three of his friends (out of 10) for dinner, then we know that he has to choose three amongst his 10 friends. This can be done in: 10_C3

The key word in the question still is: at least, this means the friend can call 3 or 4 or 5 or…10 friends over for dinner. Mathematically, this can be shown as:

Therefore, the man can invite his 10 friends in [(1024 – 1)– (10+45)]or 968 ways.

Example:
A question paper has 5 multiple choice questions and 5 true or false questions. Each multiple choice question is provided with 3 choices. Find the number of ways that an examinee can attempt one or more questions of the given 10 questions.

(1)  3¹⁰ (2) 6⁵-1
(3)  8⁵ -1    (4) 12⁵ -1

Solution: 

Number of ways of dealing with each multiple choice question = 4(3 ways of attempting and one way of not attempting)
Number of ways of dealing with each true or false question = 3 (2 ways of attempting and one way of not attempting)
Therefore Number of ways of attempting one or more questions = (4⁵ × 3⁵) – 1 = 12⁵ – 1

Read concept of Permutations and Combinations here.