Permutation, Combination and Probability (UPSC) Part-II

Total Questions: 51

31. The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible? [2022-II]

Correct Answer: (c) 24
Solution:

When A or E are first or fourth letter then arrangement = 2! × 3! = 12
When A or E are second or fifth letter then aggangement = 2! × 3! = 12
Hence, total arrangement = 12 + 12 = 24

32. A, B and C are three places such that there are three different roads from A to B. four different roads from B to C and three different roads from A to C. In how many different ways can one travel from A to C using these roads? [2022-II]

Correct Answer: (c) 15
Solution:

Number of different way from A to C =3 + 3 × 4 = 15

33. There is a numeric lock which has a 3 digit PIN. The PIN contains digits 1 to 7. There is no repetition of digits. The digits in the PIN from left to right are in decreasing order. Any two digits in the PIN differ by at least 2. How many maximum attempts does one need to find out the PIN with certainty? [2022-II]

Correct Answer: (c) 10
Solution:

Hence, maximum number of attempts required = 6 + 3 + 1 = 10

34. There are eight equidistant points on a circle. How many right-angled triangles can be drawn using these points as vertices and taking the diameter as one side of the triangle? [2022-II]

Correct Answer: (a) 24
Solution:Total number of diameters that can be drawn from 8 points on the circle = 8/2 = 4
Now, Numver of right angle triangle from n diagonal
= 2 × ⁿC₂ = 2n(n – 1)
Here, n = 4
∴ Number of diagonal = 2 × 4 (4 – 1) = 24

35. One non-zero digit, one vowel and one consonant from English-alphabet (in capital) are to be used in forming passwords, such that each password has to start with a vowel and end with a consonant. How many such passwords can be generated? [2022-II]

Correct Answer: (c) 945
Solution:

Hence, total different password = 5 × 9 × 21 = 945

36. There are 9 cups placed on a table arranged in equal number of rows and columns out of which 6 cups contain coffee and 3 cups contain tea. In how many ways can they be arranged so that each row should contain at least one cup of coffee? [2022-II]

Correct Answer: (d) 81
Solution:

∴ Number of arrangement = 3 × 3 = 9
Again cups in 3 row arrange in 3! = 6 way
∴ Number of ways = 9 × 6 = 54
And cups in 3 row are arrange in 3 × 3 × 3 = 27 ways
Hence, total number of ways = 54 + 27 = 81

37. Raj has ten pairs of red, nine pairs of white and eight pairs of black shoes in a box. If he randomly picks shoes one by one (without replacement) from the box to get a red pair of shoes to wear, what is the maximum number of attempts he has to make? [2023-II]

Correct Answer: (d) 45
Solution:Worst case analysis:
In this case Raj would be completing a red pair in the end. So, let’s say Raj drowns 18 white and 16 black shoes first and then 9 red shoes (all of left foot).

38. In how many way can a batsman score exactly 25 runs by scoring single runs, fours and sixes only, irrespective of the sequence of scoring shots? [2023-II]

Correct Answer: (b) 19
Solution:

39. There are four letters and four envelopes and exactly one letter is to be put in exactly one envelop with the correct address. If the letters are randomly inserted into the envelopes, then consider the following statements: [2023-II]

1. It is possible that exactly one litter goes into an incorrect envelope.
2. There are only six ways in which only two letters can go into the correct envelopes.

Which of the statements given above is/are correct?

Correct Answer: (b) 2 only
Solution:

  1. Total number of ways of putting all four letters into four envelope = 4! = 24
    Suppose A, B, C and D are 4 envelope and L1, L2, L3 and L4 are four letters with correct address respectively.
    S1: When L1 is placed in either of envelope B, C or D into an incorrect address, then placed into envelope A into an incorrect address.
    Hence, it is not possible that exactly one letter goes into an incorrect envelope.
    Hence, Statement 1 is not correct.
    S2: Possible ways in which only two letters can go into correct address.

EnvelopeABCD
1.L₂L₁L₃L₄
2.L₃L₂L₁L₄
3.L₄L₂L₃L₁
4.L₁L₃L₂L₄
5.L₁L₄L₃L₂
6.L₁L₂L₄L₃

Only 6 possible ways in which only two letters can go into correct envelopes.
Hence, Statement 2 is correct.

40. How many distinct 8-digit numbers can be formed by rearranging the digits of the number 11223344 such that odd digits occupy odd positions and even digits occupy even positions? [2023-II]

Correct Answer: (c) 36
Solution:

Number of ways of arrangement of odd digits

at odd place = 4! / (2! × 2!) = 3 × 2 = 6

Number of ways of arrangement of even digits at
Even place = 4! / (2! × 2!) = 3 × 2 = 6.

Total Arrangement = 6 × 6 = 36