# Permutations & Combinations Mcqs - Set 4

1)   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?

a. 36
b. 81
c. 91
d. 116
 Answer  Explanation ANSWER: 91 Explanation: 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.

2)   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?

a. 12
b. 48
c. 36
d. 42
 Answer  Explanation ANSWER: 42 Explanation: 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.

3)   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?

a. 19
b. 120
c. 624
d. 1024
 Answer  Explanation ANSWER: 624 Explanation: 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.

4)   In how many different ways can six players be arranged in a line such that two of them, Asim and Raheem are never together?

a. 120
b. 240
c. 360
d. 480
 Answer  Explanation ANSWER: 480 Explanation: 1. As there are six players, So total ways in which they can be arranged = 6!ways =720. A number of ways in which Asim and Raheem are together = 5!x2 = 240. Therefore, Number of ways when they don’t remain together = 720 -240 =480.

5)   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?

a. 5
b. 6
c. 7
d. 8
 Answer  Explanation ANSWER: 7 Explanation: 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.

6)   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?

a. 20
b. 40
c. 512
d. 1024
 Answer  Explanation ANSWER: 1024 Explanation: Each question can be answered in 2 ways. 10 Questions can be answered = 2 power of 10= 1024 ways.

7)   Six points are marked on a straight line and five points are marked on another line which is parallel to the first line. How many straight lines, including the first two, can be formed with these points?

a. 29
b. 32
c. 55
d. 30
 Answer  Explanation ANSWER: 32 Explanation: We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear and no other set of three points, except those that can be selected out of these 6 points are collinear. Hence, the required number of straight lines = ¹¹C₂ – ⁶C₂ – ⁵C₂ + 1 + 1 = 55 – 15 – 10 + 2 = 32

8)   A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected if it should have at least one senior?

a. ²²C₁₀
b. ²²C₁₀ + 1
c. ²²C₉ + ¹⁰C₁
d. ²²C₁₀ – 1
 Answer  Explanation ANSWER: ²²C₁₀ – 1 Explanation: The total number of ways of forming the group of ten representatives is ²²C₁₀. The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way The required number of ways = ²²C₁₀ – 1

9)   A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected, if it should have 5 seniors and 5 juniors?

a. ¹²C₅ * 10
b. ¹²C₇ * 10
c. ¹²C₇ * ¹⁰C₅
d. 12 * ¹⁰C₅
 Answer  Explanation ANSWER: ¹²C₇ * ¹⁰C₅ Explanation: Here, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅ = ¹²C₅ = ¹²C₇

10)   A group consists of 4 men, 6 women and 5 children. In how many ways can 3 men, 2 women and 3 children selected from the given group?

a. 300
b. 450
c. 600
d. 750
 Answer  Explanation ANSWER: 600 Explanation: The number of ways of selecting three men, two women and three children is: = ⁴C₃ * ⁶C₂ * ⁵C₃ = (4 * 3 * 2)/(3 * 2 * 1) * (6 * 5)/(2 * 1) * (5 * 4 * 3)/(3 * 2 * 1) = 4 * 15 * 10 = 600 ways.

11)   A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group?

a. 300
b. 600
c. 750
d. 900
 Answer  Explanation ANSWER: 600 Explanation: Two men, three women and one child can be selected in ⁴C₂ * ⁶C₃ * ⁵C₁ ways = (4 * 3)/(2 * 1) * (6 * 5 * 4)/(3 * 2) * 5 = 600 ways.

12)   Find the number of ways of arranging the letters of the word “MATERIAL” such that all the vowels in the word are to come together?

a. 720
b. 1440
c. 1860
d. 2160
 Answer  Explanation ANSWER: 1440 Explanation: In the word, “MATERIAL” there are three vowels A, I, E. If all the vowels are together, the arrangement is MTRL’AAEI’. Consider AAEI as one unit. The arrangement is as follows. M T R L A A E I The above 5 items can be arranged in 5! ways and AAEI can be arranged among themselves in 4!/2! ways. Number of required ways of arranging the above letters = 5! * 4!/2! = (120 * 24)/2 = 1440 ways.

13)   The number of new words that can be formed by rearranging the letters of the word ‘ALIVE’ is__________?

a. 24
b. 23
c. 119
d. 120
 Answer  Explanation ANSWER: 119 Explanation: Number of words which can be formed = 5! – 1 = 120 – 1 = 119.

14)   A delegation of 5 members has to be formed from 3 ladies and 5 gentlemen. In how many ways the delegation can be formed, if 2 particular ladies are always included in the delegation?

a. 20
b. 54
c. 42
d. 60
 Answer  Explanation ANSWER: 20 Explanation: There are three ladies and five gentlemen and a committee of 5 members to be formed. Number of ways such that two ladies are always included in the committee = ⁶C₃ = (6 * 5 * 4)/6 = 20.

15)   The number of sequences in which 7 players can throw a ball, so that the youngest player may not be the last is -.

a. 4000
b. 2160
c. 4320
d. 5300
 Answer  Explanation ANSWER: 4320 Explanation: x Not younger_______ ↑ The last ball can be thrown by any of the remaining 6 players. The first 6 players can throw the ball in ⁶P₆ ways. The required number of ways = 6(6!) = 4320

16)   In how many ways can live boys and three girls sit in a row such that all boys sit together?

a. 4800
b. 5760
c. 2880
d. 15000
 Answer  Explanation ANSWER: 2880 Explanation: Treat all boys as one unit. Now there are four students and they can be arranged in 4! ways. Again five boys can be arranged among themselves in 5! ways. Required number of arrangements = 4! * 5! = 24 * 120 = 2880.