## Wipro Sample Combinations Questions

Below are three aptitude questions requiring you to employ combinations in equations to arrive at results.

Question 1

Mr. Krishnaraj Manickam, a final year student of B. Tech -Information Technology from a College in Kanpur appeared for a recruitment test conducted by a leading IT company in campus interview. The question consisted of two parts – Part A and Part B. There were six questions in each part. A candidate has to answer seven questions in all choosing not more than five from one part. In how many ways a candidate can attempt this paper assuming he answers six questions?

a) 684 b) 780 c)612 d) none of these.

Solution :

Each section has six questions. Candidate has to answer totally seven questions, choosing not more than 5 from each part.
Possible options are:
4 questions from Part A AND 3 questions from Part B OR 3 questions from Part A AND 4 questions from Part B OR 5 questions from Part A AND 2 questions from Part B OR 2 questions from Part A AND 5 questions from Part B.

Note : i) Number of ways to choose r questions from among n questions can be obtained by the 'combinations' formula nCr. ii) To solve questions like those we are seeing currently, all the ANDs should be replaced by 'X' (multiplication symbol) and ORs should be replaced by '+' (addition symbol)

Based on above two points in 'Notes' , Applying the combinations formula, replacing ANDs with x and ORs with + we get
Total Possible Ways Of Choosing 6 Questions = 6C4 x 6C3 + 6C3 x 6C4 + 6 C5 x 6C2 + 6C2 x 6C5 = 300 + 300 + 90 + 90 = 780

Question 2

SSCT Consultancy Services wanted to select 78 candidates from the five engineering Colleges in Andhra Pradesh, Tamil Nadu, Kerala and Karnataka. They wanted only those candidates who have cleared all papers in semester exams to appear for their recruitment test. The test paper set by them consisted of two parts--- Part I and Part II. Each part consisted of 7 questions and the candidates are supposed to answer only six questions. Candidates were given the freedom to choose any 7 questions across Part I and Part II but with a simple condition that at least one question should be answered from part I. In how many ways a candidate can attempt this paper assuming he answers six questions?

a) 3295 b) 2996 c) 2955 d) none of these.

Solution :

Let us discuss possible ways for any given candidate to choose questions provided he answers at least one question from part A.
1 question from part A AND 5 questions from part B OR 2 questions from part A AND 4 questions from part B OR 3 questions from part A AND 3 questions from part B OR
4 questions from part A AND 2 questions from part B OR 5 questions from part A AND 1 question from part B OR 6 questions from part A AND none from part B

For above combinations possible ways of choosing questions
= 7C1 X 7C5 + 7C2 X 7C4 + 7C3 X 7C3 + 7C4 X 7C2 + 7C5 X 7C1 + 7C6 X 7C0
= 7 x 21 + 21 x 35 + 35 x 35 + 35 x 21 + 21 x 7 + 7
= 147 + 735 + 1225 + 735 + 147 + 7 = 2996

Question 3

Raja Ganapathy, a final year student of Emperor Engineering College was aspiring to join a computer company as a software engineer had to attend a test. The test paper consisted of technical questions put in two parts A and part B. Each part consisted of 5 questions. Across Part A and Part B he has to answer 6 questions in total with a condition that he has to answer at least 2 questions from each of the sections. However, after reading the questions he realized he knew answers for 4 questions from Part A and 3 questions from Part B. In how many ways Raja Ganapathy can attend this technical test paper?

a) 100 b) 200 c) 50 d) none of these.

Solution :

He knew at most 4 questions from Part A and 3 from Part B. Also there is a condition that he has to answer at least 2 questions from each of the sections. Based on these constraints he can choose his 6 questions as below :
Note : He knews at most only 3 questions from Part B (as per question). Therefore, though rules permit he will be unable to answer 4 questions from Part B.

4 Questions from Part A AND 2 Questions from Part B OR 3 Questions from Part A AND 3 Questions from Part B
For above combinations possible ways of choosing questions
= 5C4 X 5C2 + 5C3 X 5C3
= 5 X 10 + 10 X 10 = 50 + 100 = 150

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