Permutation, Combination and Probability (UPSC) Part-II

Total Questions: 51

21. How many five-digit prime numbers can be obtained by using all the digits 1, 2, 3, 4 and 5 without repetition of digits? [2020-II]

Correct Answer: (a) Zero
Solution:
  1. Sum of digits = 1 + 2 + 3 + 4 + 5 = 15.

and 15 is divisible by 3. Thus all 5 digits numbers obtained by the digits 1, 2, 3, 4 and 5 will be divisible by 3.
All 5 digits numbers obtained by using all the digits 1, 2, 3, 4 and 5 without repetition of digits, will be divisible by 3.
Hence, no. 5 digits prime number can be obtained by the digits 1, 2, 3, 4 and 5.

22. How many different 5-letter words (with or without meaning) can be constructed using all the letters of the word 'DELHI' so that each word has to start with D and end with I? [2020-II]

Correct Answer: (d) 6
Solution:First letter of each word is D and last letter of each work is I.
So, number of ways of arrange first and last letter of the word = 1
Remaining letter = 3.
Remaining 3 letters arrange at 3 positions in
³P₃ = 3! = 3×2×1 = 6 ways.
Hence, total number of required word = 6×1 = 6.
And that words are DELHI, DLHEI, DHELI, DEHLI, DLEHI, DHLEI.

23. How many different sums can be formed with the denominations ₹50, ₹100, ₹200, ₹500 and ₹2,000 taking at least three denominations at a time? [2020-II]

Correct Answer: (a) 16
Solution:Different sums that can be obtained by taking at least 3 out of 5 denominations are -
⁵C₁ + ⁵C₄ + ⁵C₅ = 10 + 5 + 1 = 16.

24. Consider all 3-digit numbers (without repetition of digits) obtained using three non-zero digits which are multiples of 3. Let S be their sum. [2021-II]

Which of the following is/are correct?
1. S is always divisible by 74.
2. S is always divisible by 9.

Select the correct answer using the code given below:

Correct Answer: (c) Both 1 and 2
Solution:3 digit numbers obtained using digit of 3 i.e. 3, 6 and 9.
So, numbers are 369, 369, 639, 693, 936 and 963.
So, sum S = 369 + 396 + 639 + 693 + 636 + 963 = 3996
Here, number 3996 is divisible by 74 and 9.

25. On a chess board, in how many different ways can 6 consecutive squares be chosen on the diagonals along a straight path? [2021-II]

Correct Answer: (b) 6
Solution:Here's a typical 8 × 8 chess board:

There are two diagonals, each having 8 squares.

On one diagonal, 6 consecutive squares can be chosen in 3 ways. So, total number of ways of choosing 6 consecutive squares on the diagonals along a straight path = 3 + 3 = 6

26. Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed? [2021-II]

Correct Answer: (b) 6
Solution:

Hence, total number of ways = 4!/2!2! = 6

27. There are 6 persons arranged in a row. Another person has to shake hands with 3 of them so that he should not shake hands with two consecutive persons. In how many distinct possible combinations can the handshakes take place? [2021-II]

Correct Answer: (b) 6
Solution:Total number of persons in the row = 6
Possible combination for hand shake = ⁶C₂ = 6×5/2 = 4

28. The digits 1 to 9 are arranged in three rows in such a way that each row contains three digits, and the number formed in the second row is twice the number formed in the first row; and the number formed in the third row is thrice the number formed in the first row. Repetition of digits is not allowed. If only three of the four digits 2, 3, 7 and 9 are allowed to use in the first row, how many such combinations are possible to be arranged in the three rows? [2022-II]

Correct Answer: (c) 2
Solution:
  1. Here all numbers are of 3 digits, then number in first row must be less than 333 and number in second row must be than 666.
    Now, among the given digits 2, 3, 7 and 9.
    Number in first row would be either
    237 or 239 or 273 or 327 or 329

Possibility-IPossibility-IIPossibility-IIIPossibility-IVPossibility-V
First row237239273329327
Second row474478546658654
Third row711717819987981

Here, only Possibility–I and III are valid. In all other possibility digits are repeated.

29. In a tournament of Chess having 150 entrants, a player is eliminated whenever he loses a match. It is given that no match results in a tie/draw. How many matches are played in the entire tournament? [2022-II]

Correct Answer: (c) 149
Solution:In every matches, one players is eliminated out of 150 player, only one player is winner and 150 – 1 = 149 players is eliminated
Hence, total number of games played
= Number of players eliminated in each match = 149.

30. How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5? [2022-II]

Correct Answer: (b) 12
Solution:

3-digit natural numbers have all odd different digit and also divisible by 5.
∴ Last digit of each number is 5.
First and second digits are any two digits out of four odd digit (1, 3, 7 and 9).

Hence, total number = ⁴P₂ = 4 × 3 = 12