Six identical cards are placed on a table. Each card has number '1' marked on one side and number '2' marked on its other side. All the six cards are placed in such a manner that the number '1' is on the upper side. In one try, exactly four(neither more nor less) cards are turned upside down. In how many least number of tries can the cards be turned upside down such that all the six cards show number '2' on the upper side?
UPSC-2014
This problem involves a state-change process where we need to find the minimum number of operations to reach a target state. We start with all six cards showing '1'. The goal is to have all six cards show '2'. In each try, exactly four cards are flipped. Flipping a card changes its displayed number (1 to 2, or 2 to 1). We can track the number of cards showing '1' and '2'.
Initial State: All six cards show '1'. Let's represent this as (6 ones, 0 twos).
Target State: All six cards show '2'. This is (0 ones, 6 twos).
First Try: We must flip exactly four cards. Since all cards show '1', flipping any four cards will change them to '2'.
State after 1st try: (2 ones, 4 twos).
Second Try: We need to flip four cards. Our goal is to increase the number of '2's. To do this efficiently, we should prioritize flipping '1's to '2's, and if we must, flip '2's to '1's in a way that helps us in the next step.
From (2 ones, 4 twos), we have two '1's and four '2's. We must flip four cards.
Option 1: Flip both '1's and two '2's.
State after 2nd try: (2 ones, 4 twos). This doesn't help us progress.
Option 2: Flip zero '1's and four '2's.
State after 2nd try: (6 ones, 0 twos). This brings us back to the start.
Option 3: Flip one '1' and three '2's.
State after 2nd try: (4 ones, 2 twos).
The most effective strategy is to flip as many '1's as possible to '2's, and then flip '2's to '1's if necessary to meet the 'four cards' requirement. In this case, we have two '1's. We flip these two '1's to '2's. We still need to flip \(4 - 2 = 2\) more cards. These must be chosen from the '2's, turning them into '1's.
State after 2nd try: (2 ones, 4 twos) - (2 '1's flipped to '2's) + (2 '2's flipped to '1's) = (2 ones, 4 twos). This is the same as the state after the first try. This means we are stuck in a loop if we choose this combination.
Let's re-evaluate. We have (2 ones, 4 twos). We must flip 4 cards. Let \(x\) be the number of '1's flipped and \(y\) be the number of '2's flipped. We know \(x+y=4\).
If we flip \(x\) '1's to '2's, and \(y\) '2's to '1's:
Since we want to reach (0 ones, 6 twos), we want to reduce the number of '1's. The maximum \(x\) can be is 2 (since there are only two '1's). If \(x=2\), then \(y=2\). This leads to:
This is (2 ones, 4 twos), which is the same as the previous state. This means we cannot make progress by flipping two '1's and two '2's.
What if we flip fewer '1's? If \(x=1\), then \(y=3\).
State after 2nd try: (4 ones, 2 twos).
Third Try: We are at (4 ones, 2 twos). We need to flip 4 cards. To reach (0 ones, 6 twos), we need to flip all four '1's to '2's. We have exactly four '1's.
State after 3rd try: (0 ones, 6 twos).
Thus, it takes a minimum of 3 tries to achieve the desired state.
Let's visualize the number of cards showing '1':
This sequence achieves the goal in 3 tries.
Correct Option: A) 3
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?
UPSC-2022
The problem requires us to form passwords using one non-zero digit, one vowel, and one consonant. The password must start with a vowel and end with a consonant. We need to determine the number of such unique passwords that can be generated by considering the number of choices for each position and the possible arrangements.
The password consists of three distinct characters: one non-zero digit, one vowel, and one consonant. The order is specified: it must start with a vowel and end with a consonant.
Let the password be represented by three positions: \(\text{Position 1} \quad \text{Position 2} \quad \text{Position 3}\).
The password must start with a vowel. There are 5 choices for vowels.
\[ \text{Choices for Position 1 (Vowel)} = 5 \]The password must end with a consonant. There are 21 choices for consonants.
\[ \text{Choices for Position 3 (Consonant)} = 21 \]The remaining character to be placed in Position 2 is the non-zero digit. There are 9 choices for non-zero digits.
\[ \text{Choices for Position 2 (Non-zero digit)} = 9 \]Since the choices for each position are independent, the total number of passwords is the product of the number of choices for each position.
\[ \text{Total Passwords} = (\text{Choices for Position 1}) \times (\text{Choices for Position 2}) \times (\text{Choices for Position 3}) \]\[ \text{Total Passwords} = 5 \times 9 \times 21 \]\[ \text{Total Passwords} = 45 \times 21 \]\[ \text{Total Passwords} = 945 \]C) 945
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?
UPSC-2022
The problem asks for the number of ways to arrange 9 cups (6 coffee, 3 tea) in a 3x3 grid such that each row contains at least one cup of coffee. We will first determine the number of ways to place the 3 tea cups, ensuring the condition is met. Since there are 3 rows, and each row must have at least one coffee cup, this means no row can contain all 3 tea cups. Also, no row can contain two tea cups and one coffee cup, because if one row has two tea cups, the remaining tea cup must be placed in another row, leaving one row with three coffee cups, which is allowed. The condition implies that we cannot have a row with zero coffee cups, which means a row cannot be filled entirely with tea cups. Since there are only 3 tea cups, this condition is automatically satisfied if we distribute the 3 tea cups such that no row gets all three.
Understand the arrangement: There are 9 cups arranged in a 3x3 grid. This means there are 3 rows and 3 columns. Each row has 3 positions, and each column has 3 positions.
Identify the cups: We have 6 coffee cups (C) and 3 tea cups (T).
Condition: Each row must contain at least one cup of coffee. This means that no row can be filled entirely with tea cups.
Consider the placement of tea cups: Since there are only 3 tea cups, the condition "each row must contain at least one cup of coffee" simplifies to ensuring that the 3 tea cups are not all placed in the same row. If all 3 tea cups were in one row, that row would have 0 coffee cups, violating the condition. Therefore, the 3 tea cups must be distributed among the 3 rows such that no single row gets all 3 tea cups.
Possible distributions of tea cups across rows:
Since no row can have all 3 tea cups, the tea cups must be distributed as follows:
Case 1: One tea cup in each row (1, 1, 1 distribution)
For Row 1, there are 3 positions to place 1 tea cup. This can be done in \[\binom{3}{1}\] ways.
For Row 2, there are 3 positions to place 1 tea cup. This can be done in \[\binom{3}{1}\] ways.
For Row 3, there are 3 positions to place 1 tea cup. This can be done in \[\binom{3}{1}\] ways.
The number of ways for this case is \[\binom{3}{1} \times \binom{3}{1} \times \binom{3}{1} = 3 \times 3 \times 3 = 27\] ways.
Case 2: Two tea cups in one row, one tea cup in another row, and zero tea cups in the third row (2, 1, 0 distribution)
First, choose which row will have 2 tea cups. This can be done in \[\binom{3}{1}\] ways.
Then, choose which of the remaining two rows will have 1 tea cup. This can be done in \[\binom{2}{1}\] ways.
The last row will have 0 tea cups.
Number of ways to choose the rows for this distribution: \[\binom{3}{1} \times \binom{2}{1} = 3 \times 2 = 6\] ways.
Now, for the placement within the chosen rows:
So, for each of the 6 ways to choose the rows, the number of ways to place the tea cups is \[\binom{3}{2} \times \binom{3}{1} \times \binom{3}{0} = 3 \times 3 \times 1 = 9\] ways.
The total number of ways for Case 2 is \(6 \times 9 = 54\) ways.
Re-evaluating the condition and distribution: The condition is that each row has at least one coffee cup. This means no row can contain only tea cups. Since there are only 3 tea cups in total, the only way a row could contain only tea cups is if all 3 tea cups are in that one row. This is the only scenario that violates the condition. Therefore, we need to find the total number of ways to place the 3 tea cups in the 9 positions and subtract the ways where all 3 tea cups are in the same row.
Total ways to place 3 tea cups in 9 positions:
This is choosing 3 positions out of 9 for the tea cups: \[\binom{9}{3}\]
\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84 \] So, there are 84 total ways to place the 3 tea cups.
Ways that violate the condition (all 3 tea cups in one row):
There are 3 rows. For each row, there are \[\binom{3}{3}\] ways to place 3 tea cups in that row.
Number of ways to have all 3 tea cups in Row 1: \[\binom{3}{3} = 1\]
Number of ways to have all 3 tea cups in Row 2: \[\binom{3}{3} = 1\]
Number of ways to have all 3 tea cups in Row 3: \[\binom{3}{3} = 1\]
Total ways that violate the condition = \(1 + 1 + 1 = 3\) ways.
Number of ways satisfying the condition:
Total ways - Ways that violate the condition = \(84 - 3 = 81\).
Once the tea cups are placed, the remaining 6 positions are filled with 6 coffee cups in only 1 way. So, the number of ways to place the tea cups is the final answer.
D) 81
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?
UPSC-2017
To find the total number of rectangles and squares, we need to determine how many ways we can choose two distinct horizontal lines and two distinct vertical lines. A rectangle is formed by the intersection of two distinct horizontal lines and two distinct vertical lines. A square is a special type of rectangle. The problem asks for the maximum number of rectangles and squares, which implies finding all possible rectangles.
Correct Option: C) 36
A bag contains 15 red balls and 20 black balls. Each ball is numbered either 1 or 2 or 3. 20% of the red balls are numbered 1 and 40% of them are numbered 3. Similarly, among the black balls, 45% are numbered 2 and 30% are numbered 3. A boy picks a ball at random. He wins if the ball is red and numbered 3 or if it is black and numbered 1 or 2. What are the chances of his winning?
UPSC-2018
To find the chances of winning, we need to calculate the total number of winning outcomes and divide it by the total number of possible outcomes. First, we'll determine the number of red and black balls for each number (1, 2, 3). Then, we'll identify the winning conditions and sum up the number of balls that satisfy these conditions. Finally, we'll calculate the probability.
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?
UPSC-2018
To form an equilateral triangle with vertices on the circumference of a circle, the vertices must be equally spaced. This means the arcs between any two consecutive vertices of the triangle must be equal. Since there are 24 equally spaced points on the circumference, we need to find how many sets of three points can form an equilateral triangle.
Condition for an equilateral triangle: For three points on a circle to form an equilateral triangle, they must divide the circumference into three equal arcs. If the total number of points on the circumference is \(N\), then for an equilateral triangle, the points must be separated by \(N/3\) points along the circumference.
Check divisibility: In this problem, \(N = 24\). For an equilateral triangle to be formed, \(N\) must be divisible by 3. Here, \(24\) is divisible by \(3\), so equilateral triangles can be formed.
Calculate the spacing: The points forming an equilateral triangle will be separated by \(24/3 = 8\) points. So, if we pick point 1, the next point will be point \(1+8=9\), and the third point will be point \(9+8=17\). The sequence of points would be \((1, 9, 17)\).
Identify unique triangles: We can systematically list the sets of points that form equilateral triangles:
Count the triangles: If we try to form another triangle starting with point 9, we get \((9, 17, 25)\). Since point 25 is the same as point 1 (as there are 24 points), this triangle is \((9, 17, 1)\), which is the same as the first triangle \((1, 9, 17)\). Therefore, we have found all unique equilateral triangles.
General formula: The number of equilateral triangles that can be formed from \(N\) equally spaced points on a circle is \(N/3\), provided \(N\) is divisible by 3. If \(N\) is not divisible by 3, then no equilateral triangles can be formed.
In this case, the number of equilateral triangles is \(24/3 = 8\).
C) 8
