Permutation, Combination and Probability (UPSC) Part-II

Total Questions: 51

41. A box contains 14 black balls, 20 blue balls, 26 green balls, 28 Yellow balls, 38 red balls 54 white balls. Consider the following Statements: [2023-II]

1. The smallest number n such that any n balls drawn from the box randomly must contain one full group of at least one colour is 175.
2. The smallest number m such that any m balls drawn from the box randomly must contain at least one ball of each colour is 167.

Which of the above statements is/are correct?

Correct Answer: (c) Both 1 and 2
Solution:

(c) Black ball = 14, Blue ball = 20,
Green ball = 26, Yellow ball = 28
Red ball = 38, White ball = 54
Total ball = 14 + 20 + 26 + 28 + 38 + 54 = 180

  1. Number of drawn which contain one full group of atleast one colour.
    = (13 + 16 + 25 + 27 + 37 + 54) + 1 = 175
  2. Number of drawn that contain atleast one ball of each colour
    = (54 + 38 + 28 + 26 + 20) + 1 = 167

42. What is the sum of all 4-digit numbers less than 2000 formed by the digits 1, 2, 3 and 4, where none of the digits is repeated? [2023-II]

Correct Answer: (a) 7998
Solution:Number < 2000
Then, first digit will be 1.
Last 3-digits will be 3! = 3 × 2 × 1 = 6
Sum of the numbers = 1234 + 1243 + 1324 + 1342 + 1423 + 1432 = 7998

43. What is the number of selections of 10 consecutive things out of 12 things in a circle taken in the clockwise direction? [2023-II]

Correct Answer: (c) 12
Solution:

Number of selection of 10 consecutive things out of 12 things in clockwise = 12

44. 40 children are standing in a circle and one of them (say child-1) has a ring. The ring is passed clockwise. Child-1 passes on to child-2, child-2 passes on to child-4, child-4 passes on to child-7 and so on. After how many such changes (including child-1) will the ring be in the hands of child-1 again? [2023-II]

Correct Answer: (b) 15
Solution:Child-1 Passes ring to child-2. Child-7 passes ring to child-11 and so on.

Here, child, no-46 is actually child-6, 56 is child-16, child 79 is child-39, child-121 is child-1. In 15th changes ring will be in the hands of child-1 again.

In this way Raj has drawn 44 shoes and a red pair is still not complete. The next shoe drawn will complete a red pair.

45. In an examination, the maximum marks for each of the four papers namely P, Q, R and S are 100. Marks scored by the students are in integers. A student can score 99% in n different ways. What is the value of n? [2023-II]

Correct Answer: (d) 35
Solution:

Total marks = 100 + 100 + 100 + 100 = 400
99% score = 400 × 99/100 = 396

PQRSTotal way
100100100964! / 3! = 4
10010099974! / 2! = 12
10010098984! / (2!2!) = 6
1009999984! / 2! = 12
999999994! / 4! = 1

Total number of ways = 4 + 12 + 6 + 12 + 1 = 35

46. There are five persons P, Q, R, S and T each one of whom has to be assigned one task. Neither P nor Q can be assigned Task-1. Task-2 must be assigned to either R or S. In how many ways can the assignment be done? [2023-II]

Correct Answer: (d) 24
Solution:Understanding from the question, there are two tasks - 1 and 2 and five persons - P, Q, R, S and T. As per the given constraints, the number of ways of assigning tasks to P and Q is just 1 (because P as well as Q can be given only Task-2)
No. of ways of assigning task to T is 2 (T can be given any of the two tasks)
No. of ways of assigning tasks to R and S is 3 (R-1, S-2; R-2, S-1; R-2, S-2)
So, total number of ways of assigning tasks to five persons is 1 × 2 × 3 × 4 = 24

47. What percent of water must be mixed with honey so as to gain 20% by selling the mixture at the cost price of honey? [2024-II]

Correct Answer: (a) 20%
Solution:

Let C.P. of Honey/liter is ₹100
S.P. of mixture = 100
Profit = 20%

∴ C.P. of the mixture = (100 × 100)/120 = 250/3 /litre

∴ Required % = (50/3) / (250/3) × 100 = 20%

48. What is the rightmost digit preceding the zeroes in the value of 30³⁰? [2024-II]

Correct Answer: (d) 9
Solution:

Calculation for right most digit:
(30)³⁰ = {(3×10)³⁰} = (3³)¹⁰ × (10)³⁰
= (27)¹⁰ × (10)³⁰

Periodicity of 7 is 4.
∴ (27)¹⁰ = (27)⁴ᵏ⁺² = (27)²
∴ Rightmust digit before zero = 9

49. On January 1st, 2023, a person saved ₹1. On January 2nd, 2023, he saved ₹2 more than that on the previous day. On January 3rd, 2023, he saved ₹2 more than that on the previous day and so on. At the end of which date was his total savings a perfect square, as well a perfect cube? [2024-II]

Correct Answer: (b) 8th January, 2023
Solution:
DateSaving (in ₹)
1ˢᵗ Jan1
2ⁿᵈ Jan1 + 2
3ʳᵈ Jan1 + 2 + 2 = 1 + 2 × 2
4ᵗʰ Jan1 + 2 + 2 + 2 = 1 + 2 × 3
ⁿᵗʰ Jan1 + 2 × (n – 1)

Sum = 1 + 1 + 2 × 1 + 1 + 2 × 2 + 1 + 2 × 3 + ... 1 + 2 × (n–1)
= n + 2(1 + 2 + 3 + ... (n–1))
= n + 2 × (n–1)(n)/2
= n + n² – n = n²

Number that are square as well as Cube = 64
∴ n² = 64,
n = 8

Hence, Required date on 8ᵗʰ Jan, 2023;

50. Let p and q be positive integers satisfying p < q and p + q = k. What is the smallest value of k that does not determine p and q uniquely? [2024-II]

Correct Answer: (c) 5
Solution:

Case 1: If k = 3
p + q = 3
1 + 2 = 3
Here, p and q have unique values. Hence, this is discarded.

Case 2: If k = 4
p + q = 4
1 + 3 = 4
Here, p and q have unique values. Hence, this is discarded.

Case 3: If k = 5
p + q = 5
1 + 4 = 5, 2 + 3 = 5
Thus, here p can be 1 or 2 and q can be 4 or 3. So, here p and q have multiple possible values.