Permutation And Combination(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

PERMUTATION AND COMBINATION MCQs

Permutations And Combinations

Total Questions : 444 | Page 43 of 45 pages

Question 421. Twelve people from a club, by picking lots. One of them will host a dinner for all once in a month. The number of dinners a particular member has to host in one year is________?

One
Zero
Three
Cannot be determined

Discuss Question

Answer: Option D. -> Cannot be determined
WE CANNOT PREDICTED THE NUMBER OF DINNERS THAT THE PARTICULAR NUMBER HAS TO HOST IN ONE YEAR.

Question 422. A person ordered 5 pairs of black socks and some pairs of brown socks. The price of a black pair was thrice that of a brown pair while preparing the bill the bill clerk inter-changed the number of black and brown pairs by mistake which increased the bill by 100% what was the number of pair of brown socks in the original order?

Discuss Question

Answer: Option D. -> 25
LET THE BOUGHT X PAIRS OF BROWN SOCKS AND THE PRICE OF EACH BROWN PAIR BE Y.
THEN TOTAL COST = 5X3Y+XY
CHANGED COST = 5XY+X*3Y
ACCORDING TO THE QUESTION .
5Y+3XY –(15Y+XY)/15Y+XY X 100% = 100%
=> 5Y+3XY-15Y-XY/ 15Y+XY X100% = 100%
=> 5Y+3XY = 15Y+XY+15Y+XY
=> 5Y+3XY = 30Y+2XY
=> XY= 25Y
=> X=25.
HENCE THE ORIGINAL PAIR OF BROWN SOCKS = 25.

Question 423. In how many different ways can four books A, B, C and D be arranged one above another in a vertical order such that the books A and B are never in continuous position?

Discuss Question

Answer: Option B. -> 12
THE NUMBER OF ARRANGEMENT IN WHICH A AND B ARE NOT TOGETHER
= TOTAL NUMBER OF ARRANGEMENTS
= NUMBER OF ARRANGEMENTS IN WHICH A AND B ARE TOGETHER =4!-3!X2! = 24-12 =12.

Question 424. A two number committee comprising of one male and the female number is to be constituted out of five male and 3 females. Amongst the females, Ms. A refused to be number committee in which Mr.B is taken as the number of how many different ways can the committee be constituted?

Discuss Question

Answer: Option C. -> 14
FOR EACH VISUALIZATION LET US NAME THE FEMALES AND FEMALES(3) MALES(5)
A B C D E F G H
SINCE A CANNOT GO WITH B.
SHE WILL MAKE TEAM WITH FOUR MALES IN FOUR WAYS AD,AF,AG,AH.
SINCE THERE IS NO COMPULSION WITH FEMALES C AND E.
THEY CAN MAKE TEAM WITH 5 MALES IN 5 DIFFERENT WAYS EACH.
THEREFORE, TOTAL NUMBER OF WAYS = 4+5+5 =14

Question 425. Three flags each of different colours are available for a military exercise, Using these flags different codes can be generated by waving
I. Single flag of different colours
II. Any two flags in a different sequence of colours.
III. three flags in a different sequence of colours.
The maximum number of codes that can be generated is.

Discuss Question

Answer: Option C. -> 15
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

Question 426. 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___________?

4
12
18
None of these

Discuss Question

Answer: Option D. -> None of these
TOTAL NUMBER OF WAYS = 2C1 . 5C2 = 2 X 5!/3!2! = 2 X 10 = 210

Question 427. 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_________?

Discuss Question

Answer: Option C. -> 7
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.

Question 428. 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?

Discuss Question

Answer: Option B. -> 36
CASE I IF LADY SITS ON THE RESERVED SEAT, THEN 2 MEN CAN OCCUPY SEATS FROM 4 VACANT SEATS IN = 4P2 = 4×3 = 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

Question 429. 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?

Discuss Question

Answer: Option D. -> 42
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 = 4×3 = 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 = 12×7 = 84.
BUT EVERY MATCH INCLUDES 2 TEAMS, SO THE ACTUAL NUMBER OF MATCHES = 84/2 = 42.

Question 430. 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?

Discuss Question

Answer: Option C. -> 91
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.