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?
UPSC-2021
A bag contains 20 balls. 8 balls are green, 7 are white and 5 are red. What is the minimum number of balls that must be picked up from the bag blindfolded (Without replacing any of it) to be assured of picking at least one ball of each colour?
UPSC-2017
To assure picking at least one ball of each colour, we need to consider the worst-case scenario. This means we should assume that we pick all balls of two colours before we pick even one ball of the third colour. Once all balls of two colours are exhausted, the next ball picked will guarantee at least one of each colour.
To guarantee at least one ball of each colour, we must assume that we pick all balls of the two most numerous colours first. This is because we are blindfolded and cannot choose specific colours.
Number of green balls picked: 8
Number of white balls picked: 7
Total balls picked so far: \(8 + 7 = 15\)
At this point, we have picked all green and all white balls. The only remaining balls in the bag are red.
The very next ball we pick (the 16th ball) must be a red ball, as all other colours have been exhausted. This 16th ball will ensure that we have at least one ball of each colour (8 green, 7 white, and 1 red).
Therefore, the minimum number of balls that must be picked to be assured of picking at least one ball of each colour is \(15 + 1 = 16\).
B) 16
The number of parallelograms that can be formed from a set of four parallel lines) intersecting another set of four parallel lines, is
UPSC-2019
A parallelogram is formed by two pairs of parallel lines. In this problem, we have two sets of parallel lines intersecting each other. To form a parallelogram, we need to select two lines from the first set and two lines from the second set.
A parallelogram is a quadrilateral with two pairs of parallel sides. When two sets of parallel lines intersect, any choice of two lines from the first set and two lines from the second set will form a parallelogram.
This is a combination problem. We need to choose 2 lines out of 4. The formula for combinations is \(C(n, k) = \frac{n!}{k!(n-k)!}\).
\[ C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6 \]So, there are 6 ways to choose two parallel lines from the first set.
Similarly, we need to choose 2 lines out of 4 from the second set.
\[ C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6 \]So, there are 6 ways to choose two parallel lines from the second set.
To form a parallelogram, we combine one choice from the first set with one choice from the second set. Therefore, the total number of parallelograms is the product of the number of ways to make each choice.
\[ \text{Total parallelograms} = C(4, 2) \times C(4, 2) = 6 \times 6 = 36 \]Correct Option: D) 36
Suppose you have sufficient amount of rupee currency in three denominations: 1, 10 and 50. In how many different ways can you pay a bill of 107?
UPSC-2019
This problem can be solved by systematically listing all possible combinations of denominations (1, 10, 50) that sum up to 107. We can start by fixing the number of 50-rupee notes and then find the combinations for 10-rupee and 1-rupee notes for the remaining amount.
Correct Option: C) 18
How many five-digit prime numbers can be obtained by using all the digits 1, 2, 3, 4 and 5 without repetition of digits?
UPSC-2020
To determine the number of five-digit prime numbers that can be formed using the digits 1, 2, 3, 4, and 5 without repetition, we need to recall a fundamental divisibility rule for prime numbers. A number is prime if it is only divisible by 1 and itself. A common divisibility rule states that if the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3. If a number is divisible by 3 and is greater than 3, it cannot be a prime number.
Correct Option: A) Zero
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?
UPSC-2020
The problem asks us to find the number of different sums that can be formed using given denominations, taking at least three denominations at a time. This is a combination problem where we need to consider sums formed by selecting 3, 4, or 5 denominations from the available 5 denominations. We must also check for duplicate sums.
Identify the denominations: The given denominations are ₹ 50, ₹ 100, ₹ 200, ₹ 500, and ₹ 2,000. Let's denote them as \(D_1 = 50\), \(D_2 = 100\), \(D_3 = 200\), \(D_4 = 500\), \(D_5 = 2000\).
Calculate sums taking exactly 3 denominations: The number of ways to choose 3 denominations from 5 is given by the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\). Here, \(n=5\) and \(k=3\).
\[ C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10 \]The sums are:All these 10 sums are distinct.
Calculate sums taking exactly 4 denominations: The number of ways to choose 4 denominations from 5 is \(C(5, 4)\).
\[ C(5, 4) = \frac{5!}{4!(5-4)!} = \frac{5!}{4!1!} = 5 \]The sums are:All these 5 sums are distinct from each other and also distinct from the sums formed using 3 denominations.
Calculate sums taking exactly 5 denominations: The number of ways to choose 5 denominations from 5 is \(C(5, 5)\).
\[ C(5, 5) = \frac{5!}{5!(5-5)!} = \frac{5!}{5!0!} = 1 \]The sum is:This sum is distinct from all previously calculated sums.
Total distinct sums: Add the number of distinct sums from each case.
\[ \text{Total sums} = (\text{sums with 3 denominations}) + (\text{sums with 4 denominations}) + (\text{sums with 5 denominations}) \]\[ \text{Total sums} = 10 + 5 + 1 = 16 \]Check for duplicate sums: Since all denominations are distinct positive integers and we are forming sums by adding different sets of these denominations, it's highly unlikely to get duplicate sums. For example, if we have a set of denominations \(\{d_1, d_2, ..., d_k\}\) and another set \(\{d'_1, d'_2, ..., d'_m\}\), and if these sets are different (meaning they don't contain the exact same denominations), their sums will generally be different, especially with the given denominations. A quick check confirms that all the sums calculated are unique. For instance, the smallest sum with 3 denominations is 350, and the largest is 2700. The smallest sum with 4 denominations is 850, and the largest is 2800. The sum with 5 denominations is 2850. There are no overlaps.
A) 16. As calculated, there are 10 sums with 3 denominations, 5 sums with 4 denominations, and 1 sum with 5 denominations. All these sums are distinct, leading to a total of \(10 + 5 + 1 = 16\) different sums.
