PERMUTATION & COMBINATION
- A selection is to be made for one post of Principal and two posts of Vice-Principal.Amongst the six candidates called for the interview, only two are eligible for the post of Principal while they all are eligible for the post of Vice-Principal.
The number of possible combinations of selectees is (2015)
(A) 4
(B) 12
(C) 18
(D) None of the above
Answer-D
For the post of Principal, there are two eligible candidates out of six. So, the number of ways to select one candidate for the post of Principal is 2C1 (combination of 2 items taken 1 at a time), which is equal to 2.
For the posts of Vice-Principal, all six candidates are eligible. Therefore, the number of ways to select two candidates for the posts of Vice-Principal is 6C2 (combination of 6 items taken 2 at a time), which is equal to 15.
Total combinations = Number of combinations for Principal * Number of combinations for Vice-Principal
Total combinations = 2 * 15 = 30
Therefore, there are 30 possible combinations of selectees for the given positions.
None of the given options (a) 4, (b) 12, or (c) 18 matches the correct answer.
Hence, the correct answer is (d) None of the above.
- A student has to opt for 2 subjects out of 5 subjects for a course, namely, Commerce, Economics, Statistics, Mathematics I and Mathematics II. Mathematics II can be offered only if Mathematics I is also opted. The number of different combinations of two subjects which can be opted is (2015)
(A) 5
(B) 6
(C) 7
(D) 8
Answer-C
To find the number of different combinations of two subjects that can be opted, we can use the concept of combinations.
Since Mathematics II can only be offered if Mathematics I is also opted, we need to consider two cases:
Case 1: Mathematics I is chosen
In this case, the student can choose one more subject out of the remaining four subjects (Commerce, Economics, Statistics, Mathematics II). So the number of combinations in this case is C(4, 1) = 4.
Case 2: Mathematics I is not chosen
In this case, the student can choose two subjects out of the remaining three subjects (Commerce, Economics, Statistics). So the number of combinations in this case is C(3, 2) = 3.
Therefore, the total number of different combinations of two subjects that can be
opted is 4 + 3 = 7.
- There are 5 tasks and 5 persons. Task-1 cannot be assigned to either person-1. or person-2. Task-2 must be assigned to either person-3 or person-4. Every person is to be assigned one task. In how many ways can the assignment be done?(2015)
(A) 6
(B) 12
(C) 24
(D) 144
Answer-C
Condition I: When Task T2 is fixed for person 3
Task 1: 2 ways (excluding person 1 and person 2)
Task 2: 1 way (fixed for person 3)
Task 3: 3 ways
Task 4: 3 ways
Task 5: 3 ways
Total ways for condition I: 2 + 1 + 3 + 3 + 3 = 12
Condition II: When Task T2 is fixed for person 4
Task 1: 2 ways (excluding person 1 and person 2)
Task 2: 1 way (fixed for person 4)
Task 3: 3 ways
Task 4: 3 ways
Task 5: 3 ways
Total ways for condition II: 2 + 1 + 3 + 3 + 3 = 12
Total number of ways for both conditions: 12 + 12 = 24
Therefore, the correct answer is (c) 24.
- In a society it is customary for friends of the same sex to hug and for friends of opposite sex to shake hands when they meet. A group of friends met at a party and there were 24 handshakes. Which one among the following numbers indicates the possible number of hugs ?(2015)
(A) 39
(B) 30
(C) 21
(D) 20
Answer-C
Chances of getting 24 hand shakes..
(1) 6 girls 4 boys
(2) 8 girls 3 boys
(3)12 girls 2 boys
(1) 6 girls and 4 boys: Each boy can shake hands with 6 girls, and each girl can shake hands with 4 boys. So, the total number of handshakes is 6 * 4 = 24. The remaining friends of the same sex can form hug pairs. There are 5 girls and 3 boys, so the number of possible hugs is (5 + 4 + 3 + 2 + 1) + (3 + 2 + 1) = 21.
(2) 8 girls and 3 boys: Each boy can shake hands with 8 girls, and each girl can shake hands with 3 boys. So, the total number of handshakes is 3 * 8 = 24. The remaining friends of the same sex can form hug pairs. There are 7 girls and 2 boys, so the number of possible hugs is (7 + 6 + 5 + 4 + 3 + 2 + 1) + (2 + 1) = 28.
(3) 12 girls and 2 boys: Each boy can shake hands with 12 girls, and each girl can shake hands with 2 boys. So, the total number of handshakes is 2 * 12 = 24. The remaining friends of the same sex can form hug pairs. There are 11 girls and 1 boy, so the number of possible hugs is (11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) + 1 = 67.
The option (1) (6 girls and 4 boys) corresponds to 21 possible hugs
Therefore, the correct answer is option (c) the possible number of hugs is 21.
- In a question paper there are five questions to be attempted and the answer to each question has two choices - True (T) or False (F). It is given that no two candidates have given the answers to the five questions in an identical sequence. For this to happen the maximum number of candidates is(2016)
(A) 10
(B) 18
(C) 26
(D) 32
Answer-D
To determine the maximum number of candidates, we need to find the number of unique sequences of answers that can be formed for the five questions.
Since each question has two choices (True or False), there are 2 possibilities for each question. Therefore, the total number of possible sequences is 2^5 = 32.
However, we are given that no two candidates have given the answers in an identical sequence. This means that each candidate must have a unique sequence of answers.
Therefore, the maximum number of candidates is equal to the total number of possible sequences, which is 32.
Hence, the correct answer is option (d) 32.
- There are 4 horizontal and 4 vertical lines, parallel and equidistant to one another on a board. What is the maximum number of rectangles and squares that can be formed?(2017)
(A) 16
(B) 24
(C) 36
(D) 42
Answer-C
To form a rectangle or a square, we need to select two vertical lines and two horizontal lines, so that they form the sides of the rectangle/square.
The number of ways to select 2 lines from 4 vertical lines is 4C2 = 6.
Similarly, the number of ways to select 2 lines from 4 horizontal lines is 4C2 = 6.
Therefore, the total number of rectangles and squares that can be formed is 6 * 6 = 36.
- There are 24 equally spaced points lying on the circumference of a circle. What is the maximum number of equilateral triangles that can be drawn by taking sets of three points as the vertices?(2018)
(A) 4
(B) 6
(C) 8
(D) 12
Answer-C
Since the complete circle is 360 degrees, each of the 24 equally spaced points on the circumference of the circle subtends an angle of 360/24 = 15 degrees at the center.
To form an equilateral triangle, we need three points that are evenly spaced and form angles of 120 degrees at the center. Since each point subtends an angle of 15 degrees, the number of equilateral triangles that can be formed is 120/15 = 8.
59.For a sports meet, a winners’ stand comprising three wooden blocks is in the following form
There are six different colours available to choose from and each of the three wooden blocks is to be painted such that no two of them have the same colour.
In how many different ways can the winners' stand be painted?
(A) 120
(B) 81
(C) 66
(D) 36
Answer-A
To paint the winners' stand with three wooden blocks, we have six different colors to choose from. We need to ensure that no two blocks have the same color.
For the first block, we have six color options to choose from.
After selecting the color for the first block, we have five color options remaining for the second block.
Similarly, after selecting the colors for the first two blocks, we have four color options remaining for the third block.
Therefore, the total number of different ways to paint the winners' stand is:
6 (options for the first block) × 5 (options for the second block) × 4 (options for the third block) = 120.
Hence, the correct answer is option (a) 120
- Suppose you have sufficient amount of rupee currency in three denominations:
Rs. 1, Rs.10 and Rs.50. In how many different ways can you pay a bill of Rs.107?(2019)
(A) 16
(B) 17
(C) 18
(D) 19
Answer-C
Let the number of currency in denominations of Rs. 1, Rs. 10 and Rs. 50 be a, b and c respectively.
So, a + 10b + 50c = 107
Now, the possible values of c could be 0, 1 and 2.
For c = 0: a + 10b = 107
In this case, b can range from 0 to 10, so in total 11 ways.
For c = 1: a + 10b = 57
In this case, b can range from 0 to 5, so in total 6 ways.
For c = 2: a + 10b = 7
In this case, y can take only 1 value i.e 0. That means there is only 1 way.
Total number of ways = 11 + 6 + 1 = 18
- How many different sums can be formed with the denominations Rs 50, Rs 100,Rs 200, Rs 500 and Rs 2,000 taking at least three denominations at a time?(2020)
(A) 16
(B) 15
(C) 14
(D) 10
Answer-A
To find the different sums that can be formed with the given denominations, taking at least three denominations at a time, we can use a combination of the denominations. Since we are taking at least three denominations, we will consider combinations of 3, 4, and 5 denominations.
Let's calculate the number of sums for each case:
Case 1: Combinations of 3 denominations
We can choose 3 denominations out of the given 5 denominations.
Number of ways to choose 3 out of 5 = 5C3 = 10
So, there are 10 different sums possible with combinations of 3 denominations.
Case 2: Combinations of 4 denominations
We can choose 4 denominations out of the given 5 denominations.
Number of ways to choose 4 out of 5 = 5C4 = 5
So, there are 5 different sums possible with combinations of 4 denominations.
Case 3: Combinations of 5 denominations
We can choose all 5 denominations together, which gives us only one sum.
Total number of different sums = Sum of sums in each case
= 10 + 5 + 1
= 16
Therefore, the correct answer is (a) 16.
- 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)
(A) 24
(B) 18
(C) 12
(D) 6
Answer-D
To construct a 5-letter word using the letters of the word "DELHI" such that each word starts with "D" and ends with "I," we can follow these steps:
1.Start with the fixed letter "D."
2.Choose one letter from the remaining letters "E," "L," and "H" to fill the second position.
3.Choose one letter from the remaining two letters to fill the third position.
4.Choose one letter from the remaining one letter to fill the fourth position.
5.End with the fixed letter "I."
Let's go through each step:
Step 1: There is only one option for the first position, which is "D."
Step 2: There are three options for the second position: "E," "L," or "H."
Step 3: There are two options for the third position: "E," "L," or "H" (one letter has already been used in step 2).
Step 4: There is one option for the fourth position: the remaining letter from step 2 and step 3.
Step 5: There is only one option for the fifth position, which is "I."
By multiplying the number of options at each step, we can calculate the total number of different 5-letter words:
1 * 3 * 2 * 1 * 1 = 6
Therefore, there are 6 different 5-letter words that can be constructed using all the letters of the word "DELHI" so that each word starts with "D" and ends with "I."
Hence, the correct option is (d) 6.
- On a chessboard, in how many different ways can 6 consecutive squares be chosen on the diagonals along a straight path ?(2021)
(A) 4
(B) 6
(C) 8
(D) 12
Answer-B
On the chessboard there are 2 diagonals with 8 squares each and we have to choose 6 consecutive squares.
Let's assume a block of 6 consecutive squares as one. So we are left with 3 places on the chess board where it can be arranged by selecting any one of the places out of three
- e. 3C1 = 3.
Similarly, for other diagonal too we will get 3 different ways
So, in total we have 6 different ways of choosing 6 consecutive squares.
- Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be
formed ?(2021)
(A) 3
(B) 6
(C) 9
(D) 12
Answer-B
To determine the number of distinct numbers greater than 30000 that can be formed using the digits 2, 2, 3, 3, 3, we need to consider the possible arrangements.
Since the number must be greater than 30000, the first digit must be 3. Therefore, the first digit is fixed, and we have:
For the second digit, we have two options: 2 or 3. Let's consider both cases:
Case 1: The second digit is 2
In this case, we have the following arrangement:
2 _ _ _
Now, we have three remaining digits: 2, 3, and 3. These three digits can be arranged in 3!/(2!1!) = 3 ways.
Case 2: The second digit is 3
In this case, we have the following arrangement:
3 _ _ _
Now, we have three remaining digits: 2, 3, and 3. These three digits can be arranged in 3!/(2!1!) = 3 ways.
Therefore, in total, we have 3 + 3 = 6 distinct numbers greater than 30000 that can be formed using the given digits.
Hence, the answer is (b) 6.
- 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)
(A) 12
(B) 18
(C) 24
(D) 36
Answer-C
To find the number of arrangements where there are exactly two letters between A and E, we can treat A and E as a single entity, denoted as (AE). Then, we have four entities to arrange: (AE), B, C, and D.
The number of arrangements of these four entities is 4! = 4 × 3 × 2 × 1 = 24.
However, within the (AE) entity, A and E can be arranged in two different ways: AE or EA.
Therefore, the total number of arrangements with exactly two letters between A and E is 24 × 2 = 48.
However, we need to divide this result by 2 to account for the cases where (AE) is flipped (AE and EA are considered the same).
Therefore, the final number of such arrangements is 48 ÷ 2 = 24.
Hence, the correct answer is (c) 24.
- 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)
(A) 10
(B) 13
(C) 15
(D) 36
Answer-C
Since there are three different roads from A to B and four different roads from B to C,
we have 3 * 4 = 12 possible paths from A to C via B.
However, we also need to account for the three different roads directly from A to C.
So the total number of different ways to travel from A to C using these roads is 12 + 3 = 15.
Therefore, there are 15 different ways to travel from A to C using the given roads.
- 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)
(A) 6
(B) 8
(C) 10
(D) 12
Answer-C
All Possible Values for a 3-digit PIN are 753, 752, 751, 742, 741, 731, 642, 641, 631, 531. So, the maximum number of attempts is 10
- 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)
(A) 24
(B) 16
(C) 12
(D) 8
Answer-A
To form a right-angled triangle, one of the vertices must be the right angle. In this case, since the diameter of the circle is one side of the triangle, the right angle must be formed by two of the points that are endpoints of the diameter.
To count the number of right-angled triangles, we need to determine how many possible combinations of two points we can form using the eight equidistant points on the circle.
Choosing one point as an endpoint of the diameter leaves us with six remaining points to choose from for the second endpoint of the diameter. So there are 8 possible choices for the first point and 6 possible choices for the second point.
However, we need to divide by 2 to account for the fact that we are double-counting each triangle since the order of the points doesn't matter (choosing point A as the first endpoint and point B as the second is the same as choosing B as the first and A as the second).
Therefore, the total number of right-angled triangles that can be formed is (8 x 6) / 2 = 48 / 2 = 24.
Hence, the correct answer is (a) 24.
- One non-zero digit, one vowel and one consonant from the 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)
(A) 105
(B) 525
(C) 945
(D) 1050
Answer-C
To determine the number of passwords that can be generated, we need to consider the available choices for each position in the password.
For the first position (starting with a vowel), there are 5 options available: A, E, I, O, U (5 vowels in the English alphabet).
For the second position (any non-zero digit), there are 9 options available: 1, 2, 3, 4, 5, 6, 7, 8, 9 (excluding zero).
For the third position (ending with a consonant), there are 21 options available: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Y, Z (21 consonants in the English alphabet).
Since each position has independent choices, we can multiply the number of options together to find the total number of passwords:
Total number of passwords = Number of options for position 1 × Number of options for position 2 × Number of options for position 3
Total number of passwords = 5 × 9 × 21 = 945
Therefore, 945 passwords can be generated according to the given conditions.
- A bill for Rs 1,840 is paid in the denominations of Rs 50, Rs 20 and Rs 10 notes. 50 notes in all are used.
Consider the following statements:
- 25 notes of Rs 50 are used and the remaining are in the denominations of Rs 20 and Rs 10.
- 35 notes of Rs 20 are used and the remaining are in the denominations of Rs 50 and Rs 10.
- 20 notes of Rs 10 are used and the remaining are in the denominations of Rs 50 and Rs 20.
Which of the above statements are not correct?(2022)
(A) 1 and 2 only
(B) 2 and 3 only
(C) 1 and 3 only
(D) 1, 2 and 3
Answer-D
Let's analyze each statement and check if it is correct:
Statement 1: 25 notes of Rs 50 are used, and the remaining are in the denominations of Rs 20 and Rs 10.
To pay a bill of Rs 1,840 using only Rs 50 notes, we would need 1840/50 = 36.8 notes, which is not possible since we can't have a fraction of a note. Therefore, statement 1 is not correct.
Statement 2: 35 notes of Rs 20 are used, and the remaining are in the denominations of Rs 50 and Rs 10.
To pay a bill of Rs 1,840 using only Rs 20 notes, we would need 1840/20 = 92 notes. However, the total number of notes used is given as 50, which contradicts statement 2. Therefore, statement 2 is not correct.
Statement 3: 20 notes of Rs 10 are used, and the remaining are in the denominations of Rs 50 and Rs 20.
To pay a bill of Rs 1,840 using only Rs 10 notes, we would need 1840/10 = 184 notes. However, the total number of notes used is given as 50, which contradicts statement 3. Therefore, statement 3 is not correct.
Based on the analysis, all three statements (1, 2, and 3) are not correct.
Hence, the correct answer is (d) 1, 2, and 3.
- 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)
(A) 151
(B) 150
(C) 149
(D) 148
Answer-C
In a tournament with 150 entrants, each player needs to be eliminated to determine the winner. Since each elimination happens when a player loses a match, there will be 149 matches played to eliminate 149 players.
However, the winner is determined without needing to play another match because the last remaining player is declared the winner. Therefore, the total number of matches played in the entire tournament is 149.
Hence, the correct answer is (c) 149.
- 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)
(A) 4
(B) 3
(C) 2
(D) 1
Answer-C
To solve this problem, we can start by considering the possible combinations for the first row using the digits 2, 3, 7, and 9. Since repetition of digits is not allowed, we have four options for the first digit, three options for the second digit, and two options for the third digit. This gives us a total of 4 x 3 x 2 = 24 combinations for the first row.
Now, let's consider the second row. The number formed in the second row must be twice the number formed in the first row. Since the first row has three digits, the second row must also have three digits. However, the only possible digit options we have left are 2, 3, 7, and 9. Since we are not allowed to repeat digits, there are only two possible combinations for the second row: (2, 3, 9) and (3, 7, 9).
Finally, let's consider the third row. The number formed in the third row must be thrice the number formed in the first row. Again, we have only two possible digit combinations remaining: (2, 3, 7) and (2, 3, 9).
Therefore, there are only two possible combinations for arranging the digits in the three rows: (2, 3, 7), (3, 7, 9), and (2, 3, 9), (3, 7, 9). Thus, the answer is (c) 2.
