Permutation, Combination and Probability (UPSC) Part-I

Total Questions: 51

11. A two member committee comprising of one male and one female member is to be constituted out of five males and three females. Amongst the females, Mrs. A refused to be a member of the committee in which Mr. B is taken as the member. In how many different ways can the committee be constituted? [2003]

(a) 11
(b) 12
(c) 13
(d) 14

Correct Answer: (d) 14

Solution:For each combination, let us name the females and males

Female (3)	Male (5)
A	B
C	D
E	F
G	H

Since A can’t go with B, it will make team with four males in four ways AD, AF, AG, AH. Since there is no restriction with female C and E, they may combine with 5 males in 5 different ways each.
Total number of ways = 4 + 5 + 5 = 14

12. In a question of a test paper, there are five items each under List-A and List-B. The examinees are required to match each item under List-A with its corresponding correct item under List-B. Further, it is given that [2004]

(i) no examinee has given the correct answer.
(ii) answers of no two examinees are identical.
Which is the maximum number of examinees who took this test?

(a) 24
(b) 26
(c) 119
(d) 129

Correct Answer: (c) 119

Solution:Since, answers of no. two examinees are identical, so first item in List-A can be matched with any of the 5 items in List-B.
It can be done in 5 ways. Similarly, 2nd item in List-A can be matched with any of the remaining 4 items in List-B. It can be done in 4 ways.
Continuing in the same way,
No of ways of arranging the items = 5 × 4 × 3 × 2 × 1 = 120
Now, in this arrangement there is one such arrangement, which is the correct answer.
∴ Maximum number of examinees = no. of ways of arrangement of items = 120 - 1 = 119

13. Nine different letters are to be dropped in three different letter boxes. In how many different ways can this be done? [2004]

(a) 27
(b) 3⁹
(c) 9²
(d) 3⁹ - 3

Correct Answer: (b) 3⁹

Solution:First letter can be dropped into any of the 3 boxes. It can be done in 3 ways. Similarly second letter can also be dropped into any of the 3 boxes in 3 ways and so on. Hence, total no. of ways = 3 × 3 × 3 × ........ upto 9 times = 3⁹.

14. In how many different ways can six players be arranged in a line such that two of them, Ajit and Mukherjee, are never together? [2004]

(a) 120
(b) 240
(c) 360
(d) 480

Correct Answer: (d) 480

Solution:Total no of ways of arrangement for six players = 6!
Let us take Ajit and Mukerjee as one entity.
So, now there are (6 - 2 + 1) = 5 players
These 5 players can be arranged in 5! ways and Ajit and Mukerjee can be arranged among themselves in 2! ways.

Thus, no. of ways, when Ajit and Mukerjee are always together = 5! × 2!
Hence, no. of ways when they are never together = Total no. of ways – no. of ways when they are always together
= 6! – (5! × 2!) = 6 × 5! – (5! × 2!) = 5! (6 – 2) = 480

15. Three students are picked at random from a school having a total of 1000 students. The probability that these three students will have identical date and month of their birth, is [2004]

(a) 3/1000
(b) 3/365
(c) 1/(365)²
(d) None of these

Correct Answer: (c) 1/(365)²

Solution:

For 1st student, Probability of selecting any one day as his birthday

= 365/365 = 1

Now, the remaining two students to be selected must have same day as their birthday as for the 1st student.
Probability of rest two students, having the same birthday as that of the 1st student = (1/365) × (1/365)

Hence, required probability = 1 × (1/365)² = 1/(365)²

16. On a railway route between two places A and B, there are 20 stations on the way. If 4 new stations are to be added, how many types of new tickets will be required if each ticket is issued for a one way journey? [2005]

(a) 14
(b) 48
(c) 96
(d) 108

Correct Answer: (d) 108

Solution:For (10 + A + B) = 12 stations, no of tickets required, when 4 new stations are added for one way journey
= 12 × 4 = 48
Also, each 4 new stations require (16 – 1) = 15 new tickets for one way journey.
∴ No. of tickets for 4 new stations = 15 × 4 = 60
Hence, total new tickets = 60 + 48 = 108

17. 2 men and 1 woman board a bus in which 5 seats are vacant. One of these five seats is reserved for ladies. A woman may or may not sit on the seat reserved for ladies but a man can not sit on the seat reserved for ladies. In how many different ways can the five seats occupied by these passengers? [2005]

(a) 15
(b) 36
(c) 48
(d) 60

Correct Answer: (b) 36

Solution:

There can be two cases :
(i) Lady occupies the reserved seat
(ii) Lady does not occupy the reserved seat.

(i) □□□□→Lady
Fixing one seat for the lady, 1st man can occupy any of the remaining 4 seats in four ways and the 2nd man occupy any of the remaining 3 seats in three ways.
Hence, no. of ways = 1 × 4 × 3 = 12

(ii) □□□□
Leaving the reserved seat, 1st man can occupy any of the 4 seats in four ways. 2nd man can occupy any of the remaining 3 seats in three ways. Lady can occupy any of the remaining 2 seats in two ways.
Hence, no. of ways = 4 × 3 × 2 = 24

Thus, Total no. of ways = 12 + 24 = 36

18. A square is divided into 9 identical smaller squares. Six identical balls are to be placed in these smaller squares such that each of the three rows gets at least one ball (one ball in one square only). In how many different ways can this be done? [2005]

(a) 27
(b) 36
(c) 54
(d) 81

Correct Answer: (d) 81

Solution:

Total number of ways in which 9 balls occupy any of the 6 squares = ⁹C₆ = 84

Number of ways in which one row is not filled = 3
∴ Number of ways in which at least one ball occupies each row = 84 - 3 = 81

19. There are 10 identical coins and each one of them has 'H' engraved on its one face and 'T' engraved on its other face. These 10 coins are lying on a table and each one of them has 'H' face as the upper face. In one attempt, exactly four (neither more nor less) coins can be turned upside down. What is the minimum total number of attempts in which the 'T' faces of all the 10 coins can be brought to be the upper faces? [2005]

(a) 4
(b) 7
(c) 8
(d) Not possible

Correct Answer: (a) 4

Solution:

On the first attempt four coins are overturned.
Now, six coins are left.
In the next turn, four more are overturned. Now only two would be left. We take one more from the left over two coins and any three from the previously turned ones. Finally, the leftover coin and the three coins from the presiding step which have already been turned twice can be overturned. Thus, in four attempts, one can complete the process.

20. Ten identical particles are moving randomly inside a closed box. What is the probability that at any given point of time all the ten particles will be lying in the same half of the box? [2005]

(a) 1/2
(b) 1/5
(c) 1/2⁹
(d) 2/11

Correct Answer: (c) 1/2⁹

Solution:

Probability of a particle lying in any particular half = 1/2

∴ Probability of all 10 particles lying in either 1st half or 2nd half = (1/2)¹⁰ + (1/2)¹⁰ = 2(1/2)¹⁰ = 1/2⁹

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Permutation, Combination and Probability (UPSC) Part-I

12. In a question of a test paper, there are five items each under List-A and List-B. The examinees are required to match each item under List-A with its corresponding correct item under List-B. Further, it is given that [2004]

13. Nine different letters are to be dropped in three different letter boxes. In how many different ways can this be done? [2004]

14. In how many different ways can six players be arranged in a line such that two of them, Ajit and Mukherjee, are never together? [2004]

15. Three students are picked at random from a school having a total of 1000 students. The probability that these three students will have identical date and month of their birth, is [2004]

16. On a railway route between two places A and B, there are 20 stations on the way. If 4 new stations are to be added, how many types of new tickets will be required if each ticket is issued for a one way journey? [2005]

18. A square is divided into 9 identical smaller squares. Six identical balls are to be placed in these smaller squares such that each of the three rows gets at least one ball (one ball in one square only). In how many different ways can this be done? [2005]

20. Ten identical particles are moving randomly inside a closed box. What is the probability that at any given point of time all the ten particles will be lying in the same half of the box? [2005]

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