Permutation, Combination and Probability (UPSC) Part-I

Total Questions: 51

21. Each of two women and three men is to occupy one chair out of eight chairs, each of which numbered from 1 to 8. First, women are to occupy any two chairs from those numbered 1 to 4; and then the three men would occupy any three chairs out of the remaining six chairs. What is the maximum number of different ways in which this can be done? [2006]

(a) 40
(b) 132
(c) 1440
(d) 3660

Correct Answer: (c) 1440

Solution:2 Women can occupy 2 chairs out of the first four chairs in ⁴P₂ ways. 3 men can be arranged in the remaining 6 chairs in ⁶P₃ ways.
Hence, total no. of ways = ⁴P₂ × ⁶P₃ = 1440

22. In a tournament, each of the participants was to play one match against each of the other participants. Three players fell ill after each of them had played three matches and had to leave the tournament. What was the total number of participants at the beginning, if the total number of matches played was 75? [2006]

(a) 08
(b) 10
(c) 12
(d) 15

Correct Answer: (d) 15

Solution:Let the total no. of participants be 'n' at the beginning.
Players remaining after sometime = n - 3
Now, ⁿ⁻³C₂ + (3 × 3) = 75
(n-3)!/2!(n-5)! + 9 = 75 ⇒ n² - 7n - 120 = 0
(n + 8)(n - 15) = 0
neglecting n = -8, n = 15

23. There are three parallel straight lines. Two points, 'A' and 'B', are marked on the first line; points, 'C' and 'D' are marked on the second line; and points, 'E' and 'F', are marked on the third line. Each of these 6 points can move to any position on its respective straight line. Consider the following statements: [2006]

The maximum number of triangles that can be drawn by joining these points is 18
The minimum number of triangles that can be drawn by joining these points is zero.

Which of the statement(s) given above is/are correct?

(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2

Correct Answer: (b) 2 only

Solution:Maximum number of triangles can be formed by selecting 3, 4 or 5 points out of 6 at a time.
So, maximum no. of triangles = ⁶C₃ + ⁶C₄ + ⁶C₅ which is clearly more than 18. Now, triangles formed will be minimum i.e., zero, when the points will overlap on the same line and all the points are along the same vertical line.

24. A mixed doubles tennis game is to be played between two teams (each team consists of one male and one female.) There are four married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played? [2006]

(a) 12
(b) 21
(c) 36
(d) 42

Correct Answer: (d) 42

Solution:

Married couples: MF MF MF MF
ab, cd, ef, gh
Possible teams: ad cb eb gb
af cf ed gd
ah ch eh gf

Now, team ad can play only with: cb, cg, ch, eb, eh, gb, gf, i.e. 7
The same will apply with all teams.
So, no. of total match = 12 × 7 = 84
Since every match includes 2 teams, so the No. of matches = 84/2 = 42

25. 3 digits are chosen at random from 1,2,3,4,5,6,7,8 and 9 without repeating any digit. What is the probability that their product is odd? [2006]

(a) 2/3
(b) 5/108
(c) 5/42
(d) 8/42

Correct Answer: (a) 2/3

Solution:

Let E be the event of selecting the three numbers such that their product is odd and S be the sample space.
For the product to be odd, 3 numbers chosen must be odd.
∴ n(E) = ⁵C₃, ⇒ n(S) = ⁹C₃
∴ P(E) = n(E)/n(S) = ⁵C₃/⁹C₃ = 5/42

26. In a question paper, there are four multiple choice type questions. Each question has five choices with only one choice for its correct answer. What is the total number of ways in which a candidate will not get all the four answers correct? [2006]

(a) 19
(b) 120
(c) 624
(d) 1024

Correct Answer: (c) 624

Solution:Since, every question has five options, so no. of choices for each question = 5
∴ total no. of choices = 5 × 5 × 5 × 5 × 5 = 625
Now, no. of choices of all correct answer = 1
Hence, no. of choices for all the four answers not correct
= total no. of choices - no. of choices of all correct answer
= 625 - 1 = 624

27. Each of eight identical balls is to be placed in the squares shown in the figures given below in a horizontal direction such that one horizontal row contains six balls and the other horizontal row contains two balls. In how many maximum different ways can this be done? [2006]

(a) 38
(b) 28
(c) 16
(d) 14

Correct Answer: (a) 38

Solution:

There can be two cases:
Case (I): When x row contains 6 balls:
Then the 2 balls can be arranged in y row in ⁶P₂ ways = 15
or the 2 balls can be arranged in any of the 4 two box row in 4 ways.
So, no. of ways, when x contains 6 balls = 15 + 4 = 19.

Case (II): Similarly, no. of ways, when y row contains 6 balls = 19

As, either of case (I) or case (II) is possible.
Hence, total no. of ways = 19 + 19 = 38

28. Each of the 3 persons is to be given some identical items such that product of the numbers of items received by each of the three persons is equal to 30. In how many maximum different ways can this distribution be done? [2007]

(a) 21
(b) 24
(c) 27
(d) 33

Correct Answer: (c) 27

Solution:

Suppose three people have been given a, b and c number of items.
Then, a × b × c = 30
Now, There can be 5 cases:
Case I: When one of them is given 30 items and rest two 1 item each.

So, number of ways for (30 × 1 × 1) =

(As two of them have same number of items)
Case II : Similarly, number of ways for (10 × 3 × 1) = 3! = 6
Case III : Number of ways for (15 × 2 × 1) = 3! = 6
Case IV : Number of ways for (6 × 5 × 1) = 3! = 6
Case V : Number of ways for (5 × 3 × 2) = 3! = 6

Here, either of these 5 cases are possible.
Hence, total number of ways = 3 + 6 + 6 + 6 + 6 = 27

29. In the figure shown below, what is the maximum number of different ways in which 8 identical balls can be placed in the small triangles 1, 2, 3 and 4 such that each triangle contains at least one ball? [2007]

(a) 32
(b) 35
(c) 44
(d) 56

Correct Answer: (b) 35

Solution:

There can be five cases :
Case I : First triangle can have 5 balls and rest three 1 each.

So, number of ways for (5, 1, 1, 1) =

(∵ Three triangles are having same number of balls)

Case II : Number of ways for (4, 2, 1, 1) =

$4!2!=12\frac{4!}{2!} = 12$

(∵ Two triangles are having same number of balls)

Case III : Similarly, number of ways for (2, 2, 2, 2) =

Case IV : Number of ways for (3, 3, 1, 1) =

Case V : Number of ways for (3, 2, 2, 1) =

As, either of these five cases are possible,
Hence total number of ways = 4 + 12 + 1 + 6 + 12 = 35

30. Amit has five friends: 3 girls and 2 boys. Amit’s wife also has 5 friends: 3 boys and 2 girls. In how many maximum number of different ways can they invite 2 boys and 2 girls such that two of them are Amit’s friends and two are his wife’s? [2007]

(a) 24
(b) 38
(c) 46
(d) 58

Correct Answer: (c) 46

Solution:

Amit	Wife
(I) 1 Boy and 1 Girl	1 Boy and 1 Girl
(II) 2 Girls	2 Boys
(III) 2 Boys	2 Girls

Case I : number of ways = ²C₁ × ³C₁ × ³C₁ × ²C₁ = 36
Case II : number of ways = ³C₂ × ³C₁ = 9
Case III : number of ways = ²C₂ × ²C₂ = 1

Hence, total number of ways = 36 + 9 + 1 = 46

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Permutation, Combination and Probability (UPSC) Part-I

24. A mixed doubles tennis game is to be played between two teams (each team consists of one male and one female.) There are four married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played? [2006]

25. 3 digits are chosen at random from 1,2,3,4,5,6,7,8 and 9 without repeating any digit. What is the probability that their product is odd? [2006]

26. In a question paper, there are four multiple choice type questions. Each question has five choices with only one choice for its correct answer. What is the total number of ways in which a candidate will not get all the four answers correct? [2006]

27. Each of eight identical balls is to be placed in the squares shown in the figures given below in a horizontal direction such that one horizontal row contains six balls and the other horizontal row contains two balls. In how many maximum different ways can this be done? [2006]

28. Each of the 3 persons is to be given some identical items such that product of the numbers of items received by each of the three persons is equal to 30. In how many maximum different ways can this distribution be done? [2007]

29. In the figure shown below, what is the maximum number of different ways in which 8 identical balls can be placed in the small triangles 1, 2, 3 and 4 such that each triangle contains at least one ball? [2007]

30. Amit has five friends: 3 girls and 2 boys. Amit’s wife also has 5 friends: 3 boys and 2 girls. In how many maximum number of different ways can they invite 2 boys and 2 girls such that two of them are Amit’s friends and two are his wife’s? [2007]

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