Permutations & Combinations Aptitude Questions

Q9. A student has to opt for 2 subjects out of 5 subjects for a course. Namely commerce, economics, statistics, mathematics 1 and Mathematics 2, Mathematics 2 can be offered only if mathematics 1 has also opted. The number of different combinations of two subjects which can be opted is

Number of ways of opting a subject other than Mathematics II = 4C2. = 4x3x2!/2!x2 = 6.

Number of ways of selection of Mathematics II = 1

Therefore, Total Number of ways = 6+1 =7.

Q10. A selection is to be made for one post of principal and two posts of vice-principal amongst the six candidates called for the interview only two are eligible for the post of principal while they all are eligible for the post of vice-principal. The number of possible combinations of selectees is

Total number of ways = 2C1 . 5C2 = 2 x 5!/3!2! = 2 x 10 = 210

Q1. A question paper had 10 questions. Each question could only be answered as true(T) of False(F). Each candidate answered all the questions, Yet no two candidates wrote the answers in an identical sequence. How many different sequences of answers are possible?

Each question can be answered in 2 ways.

10 Questions can be answered = 210= 1024 ways.

Q2. Groups each containing 3 boys are to be formed out of 5 boys. A, B, C, D and E such that no group can contain both C and D together. What is the maximum number of such different groups?

Maximum number of such different groups = ABC, ABD,ABE, BCE,BDE,CEA,DEA =7.

Alternate method:

Total number of way in which 3 boys can be selected out of 5 is 5C3

Number of ways in which CD comes together = 3 (CDA,CDB,CDE)

Therefore, Required number of ways = 5C3 -3

= 10-3 =7.

Q3. In how many different ways can six players be arranged in a line such that two of them, Ajeet and Mukherjee are never together?

As there are six players, So total ways in which they can be arranged = 6!ways =720.

A number of ways in which Ajeet and Mukherjee are together = 5!x2 = 240.

Therefore, Number of ways when they don’t remain together = 720 -240 =480.

Q4. In a question paper, there are four multiple choice type questions, each question has five choices with only one choice for it’s correct answer. What is the total number of ways in which a candidate will not get all the four answers correct?

Multiple choice type questions = 1 2 3 4

Total number of ways = 5x5x5x5 =625.

A number of correct answer = 1.

Number of false answers = 625-1 =624.

5. A mixed doubles tennis game is to be played two teams(each consists of one male and one female) There are four married couples. No team is to consist a husband and his wife. What is the maximum number of games that can be played?

Married couples = MF MF MF MF

AB CD EF GH

Possible teams = AD CB EB GB

AF CF ED GD

AH CH EH GF S

Since one male can be paired with 3 other female, Total teams = 4x3 = 12.

Team AD can play only with CB,CF,CH,EB,EH,GB,GF(7 teams )

Team AD cannot play with AF, AH, ED and GD

The same will apply with all teams, So number of total matches = 12x7 = 84.

But every match includes 2 teams, so the actual number of matches = 84/2 = 42.

Q6. Three dice (each having six faces with each face having one number from 1 or 6) are ralled. What is the number of possible outcomes such that atleast one dice shows the number 2?

When the dice are rolled, the number of possible outcomes = 63 = 216.

Number of possible outcomes in which 2 does not appear on any dice = 53 = 125.

Therefore, Number of possible outcomes in which at least one dice shows 2 = 216- 125 = 91.

Q7. Three flags each of different colours are available for a military exercise, Using these flags different codes can be generated by waving

This type of question becomes very easy when we assume three colour are red(R) blue(B) and Green(G).

We can choose any colour.

Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, then number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colours, then number of ways are six i.e.., RBG, BGR, GBR, RGB, BRG, GRB.

Hence total number of ways by changing flag = 3+ 6 +6 = 15

Q8. 2 men and 1 woman board a bus of which 5 seats are vacant, one of these 5 seats is reserved for ladies. A woman may or may not sit on the seat reserved for ladies, In how many different ways can the five seats be occupied by these passengers?

Case I if lady sits on the reserved seat, then 2 men can occupy seats from 4 vacant seats in = 4P2 = 4x3 = 12ways

Case II if lady does not sit on reversed seat, then I. Woman can occupy a seat from four seats in 4 ways. I. man can occupy a seat from 3 seats in 3 ways, also I. man left can occupy a seat from remaining two seats in 2 ways.

Therefore, Total ways = 4x3x2 = 24ways

From case I and case II

Total number of ways = 12+24 = 36