Samacheer Kalvi Class 11 Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Exercise 4.5

Choose The Correct Or The Most Suitable Answer:

 

Question 1. The sum if the digits at the 10th place of all numbers formed with the help of 2, 4, 5, 7 taken all at a time is
(a) 108
(b) 36
(c) 18
Answer: (a) 108

Answer: To find the sum of the digits at the tenth place, we first figure out how often each digit appears in that specific position. Using the digits 2, 4, 5, and 7, we can create a total of \( 4! = 24 \) unique four-digit numbers. Because there are four possible positions (thousands, hundreds, tens, and units), each digit will show up in the tens place an equal number of times. This means each digit appears \( \frac{24}{4} = 6 \) times in the tens place. The sum of the given digits is \( 2 + 4 + 5 + 7 = 18 \). Therefore, the total sum of all digits found in the tens place across all these numbers is \( 18 \times 6 = 108 \). This method simplifies calculating place-value sums in permutations.
In simple words: First, we find out how many times each number (2, 4, 5, 7) appears in the 'tens' position across all possible numbers made with them. Each digit appears 6 times in that spot. Then, we add all the original digits together (18) and multiply this sum by how many times each digit appears in the tens place (6) to get the final answer.

🎯 Exam Tip: Remember that for permutations of distinct digits, each digit will appear an equal number of times in each place value, which simplifies finding the sum of digits at a particular position.

 

Question 2. In an examination, there are three multiple-choice questions and each question has 5 choices. The number of ways in which a student can fail to get all answer correct is
(a) 125
(b) 124
(c) 64
(d) 63
Answer: (b) 124
Answer: For each of the three multiple-choice questions, there are 5 possible choices. Since a student can pick any of these choices for each question, the total number of ways to answer all three questions is \( 5 \times 5 \times 5 = 5^3 = 125 \) ways. Out of all these ways, there is only one specific way to get all the answers correct. To find the number of ways a student can *fail* to get all answers correct, we subtract this single correct way from the total number of possible ways. So, the number of ways to get at least one answer wrong is \( 125 - 1 = 124 \).
In simple words: There are 5 choices for each of 3 questions, so 125 total ways to answer them. Only 1 way is perfectly correct. So, to find how many ways you can get at least one wrong answer, we take the total ways (125) and subtract the one way that is all correct (1), which gives 124 ways.

🎯 Exam Tip: When dealing with "fail to get all correct," calculate the total possible outcomes and subtract the single outcome where all answers are correct.

 

Question 3. The number of ways in which the following prize be given to a class of 30 boys first and second in mathematics, first and second in Physics, first in Chemistry and first in English is
(a) \( 30^4 \times 29^2 \)
(b) \( 30^2 \times 29^3 \)
(c) \( 30^2 \times 29^4 \)
(d) \( 30 \times 29^5 \)
Answer: (a) \( 30^4 \times 29^2 \)
Answer: Let's break down the prize distribution step-by-step. For first and second places in Mathematics, there are 30 choices for the first place and 29 choices for the second place, giving \( 30 \times 29 \) ways. Similarly, for first and second places in Physics, there are also \( 30 \times 29 \) ways. For the first place in Chemistry, there are 30 choices, and for the first place in English, there are also 30 choices. Since these are all independent events, we multiply the number of ways for each category together. Thus, the total number of ways to award all the prizes is \( 30 \times 29 \times 30 \times 29 \times 30 \times 30 \). This simplifies to \( 30^4 \times 29^2 \).
In simple words: We find how many ways to give prizes for each subject separately and then multiply them. For math and physics (first and second prize), it's 30 choices for first and 29 for second. For chemistry and English (first prize only), it's 30 choices each. Multiplying these gives the total number of ways.

🎯 Exam Tip: When distributing distinct prizes or positions among distinct individuals, use permutations (like \( P(n, k) \) or \( n \times (n-1) \times ... \)), not combinations, as the order of selection matters.

 

Question 4. The number of 5 digit numbers all digits of which are odd is
(a) 25
(b) \( 5^5 \)
(c) 625
Answer: (b) \( 5^5 \)
Answer: The odd digits are 1, 3, 5, 7, and 9. There are 5 such digits. Since we need to form a 5-digit number where every digit must be odd, and repetition of digits is allowed, each of the five places in the number can be filled in 5 ways. For example, the unit's place can be any of the 5 odd digits, the ten's place can be any of the 5 odd digits, and so on, up to the hundred thousands place. Therefore, the total number of 5-digit numbers that can be formed using only odd digits with repetition is \( 5 \times 5 \times 5 \times 5 \times 5 = 5^5 \) ways. This is a direct application of the fundamental principle of counting where each position has independent choices.
In simple words: We have 5 odd digits (1, 3, 5, 7, 9). To make a 5-digit number using only these odd digits, and we can repeat them, means each of the five spots in the number has 5 choices. So, we multiply 5 by itself 5 times to get the total number of possible numbers.

🎯 Exam Tip: When digits can be repeated, treat each position as an independent choice. If leading zeros are restricted, adjust the choices for the first digit accordingly.

 

Question 5. In 3 fingers, the number of ways four rings can be worn is
(a) \( 4^3 - 1 \)
(b) \( 3^4 \)
(c) 68
(d) 64
Answer: (b) \( 3^4 \)
Answer: In this problem, we have 4 distinct rings that need to be worn on 3 distinct fingers. For each ring, there are 3 choices of finger it can be placed on. Since the rings are distinct and the choice for one ring does not affect the choices for the others, we multiply the number of choices for each ring. Therefore, for the first ring, there are 3 options, for the second ring, there are 3 options, and so on for all four rings. The total number of ways to wear four rings on three fingers is \( 3 \times 3 \times 3 \times 3 = 3^4 = 81 \) ways. This is a classic example of distributing distinct items into distinct bins.
In simple words: Each of the four rings can be put on any of the three fingers. So, for each ring, we have 3 choices. Since there are 4 rings, we multiply 3 by itself 4 times (\( 3 \times 3 \times 3 \times 3 \)), which equals 81 different ways.

🎯 Exam Tip: When placing `n` distinct items into `k` distinct bins (where items can go into any bin), the number of ways is \( k^n \). Be careful not to confuse the base and exponent.

 

Question 6. If \( (n + 5) P_{n+1} = \frac{11(n-1)}{2} (n+3) P_n \) then the value of n are
(a) 6 and 7
(b) 2 and 11
(c) 2 and 6
Answer: (a) 6 and 7
Answer: We are given the equation \( (n + 5) P_{n+1} = \frac{11(n-1)}{2} (n+3) P_n \).
First, we need to express the permutation terms using the factorial formula \( {}^N P_R = \frac{N!}{(N-R)!} \).
For the left side: \( (n+5) P_{n+1} = \frac{(n+5)!}{((n+5)-(n+1))!} = \frac{(n+5)!}{(n+5-n-1)!} = \frac{(n+5)!}{4!} \)
For the right side: \( (n+3) P_n = \frac{(n+3)!}{((n+3)-n)!} = \frac{(n+3)!}{3!} \)
Now, substitute these back into the given equation:
\( \frac{(n+5)!}{4!} = \frac{11(n-1)}{2} \times \frac{(n+3)!}{3!} \)
We can expand \( (n+5)! \) as \( (n+5)(n+4)(n+3)! \) and \( 4! \) as \( 4 \times 3! \).
\( \frac{(n+5)(n+4)(n+3)!}{4 \times 3!} = \frac{11(n-1)}{2} \times \frac{(n+3)!}{3!} \)
Cancel out \( (n+3)! \) and \( 3! \) from both sides:
\( \frac{(n+5)(n+4)}{4} = \frac{11(n-1)}{2} \)
Multiply both sides by 4 to clear the denominators:
\( (n+5)(n+4) = 2 \times 11 \times (n-1) \)
\( n^2 + 4n + 5n + 20 = 22(n-1) \)
\( n^2 + 9n + 20 = 22n - 22 \)
Move all terms to one side to form a quadratic equation:
\( n^2 + 9n - 22n + 20 + 22 = 0 \)
\( n^2 - 13n + 42 = 0 \)
Factor the quadratic equation:
\( n^2 - 6n - 7n + 42 = 0 \)
\( n(n-6) - 7(n-6) = 0 \)
\( (n-7)(n-6) = 0 \)
This gives us two possible values for n:
\( n-7 = 0 \implies n=7 \)
\( n-6 = 0 \implies n=6 \)
Both values are positive integers, and they satisfy the condition that the "top" number in the permutation (N) must be greater than or equal to the "bottom" number (R). In this case, \( n+5 \ge n+1 \) is always true, and \( n+3 \ge n \) is always true. Thus, the possible values for n are 6 and 7. Solving algebraic equations involving factorials is a common technique in combinatorics.
In simple words: We change the P (permutation) parts into their factorial forms. Then, we simplify the equation by cancelling out common terms. This leads to a simple equation with 'n', which we solve by putting everything on one side to get a quadratic equation. Factoring this equation gives us the two possible values for 'n', which are 6 and 7.

🎯 Exam Tip: Always remember the definition of permutations in terms of factorials and simplify carefully. Pay attention to the conditions \( N \ge R \) for permutations to be valid.

 

Question 7. The product of r consecutive positive integers is divisible by
(a) r!
(b) (r-1)!
(c) (r+1)!
(d) \( r^r \)
Answer: (a) r!
Answer: The product of `r` consecutive positive integers is always divisible by `r!`. This is a fundamental property in combinatorics. It is directly related to the definition of combinations, \( {}^nC_r = \frac{n!}{r!(n-r)!} \), which must be an integer. The product of `r` consecutive integers can be written as \( n(n-1)...(n-r+1) \), which is \( \frac{n!}{(n-r)!} \) or \( {}^n P_r \). Since \( {}^n C_r = \frac{{}^n P_r}{r!} \), it means \( {}^n P_r = r! \times {}^n C_r \). As \( {}^n C_r \) is always an integer, \( {}^n P_r \) must be divisible by \( r! \). This property is very useful in proofs involving divisibility.
In simple words: If you multiply any 'r' whole numbers that come one after another, the result will always be perfectly divisible by 'r' factorial (r!). This is a basic rule in math about how numbers are structured.

🎯 Exam Tip: Recall the definition of combinations: \( {}^nC_r = \frac{{}^n P_r}{r!} \). Since \( {}^nC_r \) is always a whole number, it proves that \( {}^n P_r \) (the product of r consecutive integers) is divisible by \( r! \).

 

Question 8. The number of five-digit telephone numbers having at least one of their digits repeated is
(a) 90000
(b) 10000
(c) 30240
(d) 69760
Answer: (d) 69760
Answer: To find the number of five-digit telephone numbers with at least one repeated digit, we can use the principle of complementary counting. This means we calculate the total number of possible five-digit telephone numbers and subtract the number of five-digit telephone numbers where no digits are repeated.
For five-digit telephone numbers, it is generally assumed that the first digit can be zero (e.g., 01234). Since each of the five positions can be any of the 10 digits (0-9), and repetition is allowed, the total number of possible five-digit telephone numbers is \( 10 \times 10 \times 10 \times 10 \times 10 = 10^5 = 100000 \).
Next, let's find the number of five-digit telephone numbers where no digits are repeated. For the first digit, there are 10 choices (0-9). For the second digit, there are 9 remaining choices (since no repetition). For the third digit, there are 8 choices, and so on. This is a permutation of 10 items taken 5 at a time, denoted as \( {}^{10}P_5 \).
\( {}^{10}P_5 = 10 \times 9 \times 8 \times 7 \times 6 = 30240 \).
Finally, the number of five-digit telephone numbers with at least one repeated digit is:
Total numbers - Numbers with no repeated digits = \( 100000 - 30240 = 69760 \). This method is commonly used to solve "at least one" type of problems.
In simple words: First, find all possible 5-digit telephone numbers, even if digits repeat (that's 100,000). Then, find how many 5-digit numbers have *no* digits repeating (that's 30,240). If you subtract the numbers with no repeats from the total numbers, you get the numbers that *must* have at least one repeated digit.

🎯 Exam Tip: For "at least one" problems, it's often easier to calculate the total number of arrangements and subtract the arrangements where the condition (e.g., "no repeats") is *not* met.

 

Question 9. If \( (a^2 - a) C_2 = (a^2 - a) C_4 \) then the value of a is
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (b) 3
Answer: We are given the equation \( (a^2 - a) C_2 = (a^2 - a) C_4 \).
A key property of combinations states that if \( {}^nC_x = {}^nC_y \), then either \( x=y \) or \( x+y=n \).
In our case, \( n = a^2 - a \), \( x = 2 \), and \( y = 4 \).
Since \( x \neq y \) (because \( 2 \neq 4 \)), we must apply the second condition, which is \( x+y=n \).
So, we have:
\( 2 + 4 = a^2 - a \)
\( 6 = a^2 - a \)
Now, rearrange this into a standard quadratic equation:
\( a^2 - a - 6 = 0 \)
We can factor this quadratic equation:
\( a^2 - 3a + 2a - 6 = 0 \)
\( a(a-3) + 2(a-3) = 0 \)
\( (a+2)(a-3) = 0 \)
This gives us two possible values for `a`: \( a = -2 \) or \( a = 3 \).
However, for a combination \( {}^nC_r \) to be defined, `n` must be a non-negative integer and \( n \ge r \). Here, \( n = a^2 - a \).
If \( a = 3 \), then \( a^2 - a = 3^2 - 3 = 9 - 3 = 6 \). Since \( 6 \ge 4 \) (the largest `r` value), \( a=3 \) is a valid solution.
If \( a = -2 \), then \( a^2 - a = (-2)^2 - (-2) = 4 + 2 = 6 \). Since \( 6 \ge 4 \), \( a=-2 \) is also mathematically valid. However, since the options are positive integers and 'a' often represents a count or dimension, \( a=3 \) is chosen as the suitable answer. This problem demonstrates an important property of combinations.
In simple words: When two combinations with the same top number (like \( a^2-a \)) but different bottom numbers (2 and 4) are equal, it means the top number must be the sum of the bottom numbers. So, \( a^2-a \) equals \( 2+4 \), which is 6. We then solve this simple equation to find that 'a' can be 3 or -2, but since we usually look for positive values in these problems, 'a' is 3.

🎯 Exam Tip: Always remember the property \( {}^nC_x = {}^nC_y \implies x=y \) or \( x+y=n \). Also, check that the resulting 'n' value (in this case \( a^2-a \)) is valid for combinations (i.e., \( n \ge r \)).

 

Question 10. There are 10 points in a plane and 4 of them are collinear. The number of straight lines joining any two points is
(a) 45
(b) 40
(c) 39
(d) 38
Answer: (b) 40
Answer: To find the number of straight lines formed by 10 points where some are collinear, we first calculate the total number of lines if *no* points were collinear, and then adjust for the collinear points.
If all 10 points were non-collinear, the number of lines formed by joining any two points would be \( {}^{10}C_2 \).
\( {}^{10}C_2 = \frac{10 \times 9}{2 \times 1} = 45 \).
However, 4 of these points are collinear. If these 4 points were non-collinear, they would form \( {}^4C_2 \) lines.
\( {}^4C_2 = \frac{4 \times 3}{2 \times 1} = 6 \).
But since these 4 points are collinear, they actually form only *one* single straight line. Therefore, we subtract the lines that would have been formed by the collinear points (\( {}^4C_2 \)) and add back the single line they actually form.
Number of straight lines = (Total lines if non-collinear) - (Lines formed by collinear points) + (1 line formed by actual collinear points)
Number of straight lines = \( {}^{10}C_2 - {}^4C_2 + 1 \)
\( = 45 - 6 + 1 \)
\( = 39 + 1 \)
\( = 40 \).
This adjustment ensures we count the unique line formed by collinear points correctly.
In simple words: First, imagine all 10 points are separate and count all possible lines (45). Then, realize that the 4 points in a straight line would have made 6 lines if they weren't in a line, but they only make 1 line. So, we remove the 6 extra lines and add back the 1 true line they form. This gives \( 45 - 6 + 1 = 40 \) lines in total.

🎯 Exam Tip: When dealing with collinear points, remember the formula: Total lines = \( {}^nC_2 - {}^kC_2 + 1 \), where 'n' is the total points and 'k' is the number of collinear points.

 

Question 11. The number of ways in which a host lady invite for a party of 8 out of 12 people of whom two do not want to attend the party together is
(a) \( 2 \times {}^{11}C_7 + {}^{10}C_8 \)
(b) \( {}^{11}C_7 + {}^{10}C_8 \)
(c) \( {}^{12}C_8 - {}^{10}C_6 \)
(d) \( {}^{10}C_6 + 2! \)
Answer: (c) \( {}^{12}C_8 - {}^{10}C_6 \)
Answer: We want to find the number of ways to invite 8 people from a group of 12, with the condition that two specific people (let's call them A and B) do not want to attend the party together.
First, calculate the total number of ways to invite 8 people from 12 without any restrictions. This is a combination problem: \( {}^{12}C_8 = \frac{12!}{8!(12-8)!} = \frac{12!}{8!4!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495 \).
Next, calculate the number of ways where the two specific people (A and B) *do* attend the party together. If A and B attend together, they occupy 2 of the 8 spots. This means we still need to select \( 8 - 2 = 6 \) more people from the remaining \( 12 - 2 = 10 \) people. This can be done in \( {}^{10}C_6 \) ways.
\( {}^{10}C_6 = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \).
Finally, to find the number of ways where A and B *do not* attend together, we subtract the ways they attend together from the total unrestricted ways:
Number of ways = (Total ways) - (Ways where A and B attend together)
Number of ways = \( {}^{12}C_8 - {}^{10}C_6 \). This is a standard strategy for "not together" problems.
In simple words: First, find all the ways to pick 8 people from 12. Then, find all the ways where the two people who don't want to be together *are* picked together. Lastly, subtract the "together" ways from the "all possible" ways to find how many ways they are *not* together.

🎯 Exam Tip: For problems where "two specific items must not be together," calculate the total unrestricted combinations and subtract the combinations where those two specific items *are* together.

 

Question 12. The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines.
(a) 6
(b) 9
(c) 12
(d) 18
Answer: (d) 18
Answer: To form a parallelogram, you need two parallel lines from one set and two parallel lines from the other intersecting set. The problem states there are two sets of parallel lines: one with four lines and another with three lines.
From the set of four parallel lines, we need to choose 2 lines. The number of ways to do this is \( {}^4C_2 \).
\( {}^4C_2 = \frac{4 \times 3}{2 \times 1} = 6 \).
From the set of three parallel lines, we also need to choose 2 lines. The number of ways to do this is \( {}^3C_2 \).
\( {}^3C_2 = \frac{3 \times 2}{2 \times 1} = 3 \).
Since the choice of lines from one set is independent of the choice from the other set, we multiply the number of ways for each selection to get the total number of parallelograms.
Total number of parallelograms = \( {}^4C_2 \times {}^3C_2 = 6 \times 3 = 18 \). This method applies to any grid-like structure made by intersecting parallel lines.
In simple words: To make a parallelogram, you pick two lines from the first group of parallel lines and two lines from the second group. There are 6 ways to pick from the first group (4 lines) and 3 ways to pick from the second group (3 lines). Multiply these two numbers to find the total number of parallelograms you can make.

🎯 Exam Tip: To count parallelograms formed by intersecting parallel lines, remember to select 2 lines from each set of parallel lines and multiply the number of combinations.

 

Question 13. Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is
(a) 11
(b) 12
(c) 10
(d) 6
Answer: (b) 12
Answer: Let 'n' be the number of persons in the room. When everybody shakes hands with everybody else, each handshake involves two distinct persons. The order in which two people shake hands does not matter (person A shaking person B's hand is the same as person B shaking person A's hand). Therefore, the number of handshakes is a combination problem: \( {}^nC_2 \).
The formula for \( {}^nC_2 \) is \( \frac{n(n-1)}{2} \).
We are given that the total number of handshakes is 66. So, we set up the equation:
\( \frac{n(n-1)}{2} = 66 \)
Multiply both sides by 2:
\( n(n-1) = 132 \)
Expand the left side and rearrange into a quadratic equation:
\( n^2 - n = 132 \)
\( n^2 - n - 132 = 0 \)
Now, factor the quadratic equation. We need two numbers that multiply to -132 and add to -1. These numbers are -12 and 11.
\( (n-12)(n+11) = 0 \)
This gives two possible values for 'n':
\( n-12 = 0 \implies n = 12 \)
\( n+11 = 0 \implies n = -11 \)
Since the number of persons cannot be negative, we discard \( n = -11 \). Thus, there are 12 persons in the room. This problem illustrates a common application of combinations in real-world scenarios.
In simple words: If there are 'n' people, the number of handshakes is found by choosing any 2 people from the group, which is \( \frac{n \times (n-1)}{2} \). We know this total is 66. We set up the equation, solve it, and find that 12 people are in the room, because you can't have a negative number of people.

🎯 Exam Tip: Remember that the "handshake problem" is a classic example of combinations, specifically \( {}^nC_2 \). If given the number of handshakes, form and solve the quadratic equation to find 'n'.

 

Question 14. Number of sides of a polygon having 44 diagonals is
(a) 4
(b) 4!
(c) 11
(d) 22
Answer: (c) 11
Answer: For a polygon with 'n' sides, the total number of lines that can be drawn by joining any two vertices is \( {}^nC_2 \). These lines include both the sides of the polygon and its diagonals. Since a polygon with 'n' sides has 'n' sides (which are also lines connecting vertices), the number of diagonals is the total number of lines minus the number of sides.
Number of diagonals = \( {}^nC_2 - n \).
We are given that the number of diagonals is 44. So, we set up the equation:
\( \frac{n(n-1)}{2} - n = 44 \)
To solve for 'n', first multiply the entire equation by 2 to clear the denominator:
\( n(n-1) - 2n = 88 \)
Expand and simplify:
\( n^2 - n - 2n = 88 \)
\( n^2 - 3n = 88 \)
Rearrange into a standard quadratic equation:
\( n^2 - 3n - 88 = 0 \)
Now, factor the quadratic equation. We need two numbers that multiply to -88 and add to -3. These numbers are -11 and 8.
\( (n-11)(n+8) = 0 \)
This gives two possible values for 'n':
\( n-11 = 0 \implies n = 11 \)
\( n+8 = 0 \implies n = -8 \)
Since the number of sides of a polygon cannot be negative, we discard \( n = -8 \). Thus, the polygon has 11 sides. This formula is crucial for understanding polygon properties.
In simple words: The number of diagonals in a polygon is found by taking all possible lines between its corners (\( \frac{n \times (n-1)}{2} \)) and subtracting the lines that are actually its sides ('n'). We set this equal to 44 and solve for 'n'. After solving, we find that the polygon has 11 sides, because a polygon cannot have a negative number of sides.

🎯 Exam Tip: Remember the formula for the number of diagonals in an n-sided polygon: \( D = \frac{n(n-1)}{2} - n \). Be careful with algebraic manipulation when solving for 'n'.

 

Question 20. The number of rectangles that a chessboard has
(a) 81
(b) 99
(c) 1296
(d) 6561
Answer: (c) 1296
Number of rectangles in a chessboard is given by the product of the number of ways to choose 2 horizontal lines from 9 and 2 vertical lines from 9. Since a chessboard has 9 horizontal and 9 vertical lines, we use combinations to find the number of ways to pick two lines of each type.
\( 9C_2 \times 9C_2 \)
\( = \frac{9 \times 8}{1 \times 2} \times \frac{9 \times 8}{1 \times 2} \)
\( = 36 \times 36 \)
\( = 1296 \)
In simple words: To find how many rectangles are on a chessboard, you pick two horizontal lines and two vertical lines. Each way you pick them makes one rectangle. There are 9 horizontal lines and 9 vertical lines, so you find all possible pairs for each and multiply them together.

🎯 Exam Tip: Remember that an \( n \times n \) chessboard has \( (n+1) \) horizontal lines and \( (n+1) \) vertical lines. So, for a standard \( 8 \times 8 \) board, \( n=8 \), which means \( (8+1)=9 \) lines of each type.

 

Question 21. The number of 10 digit number that can be written by using digits 2 and 3 is
(a) \( 10C_2 + 9C_2 \)
(b) \( 2^{10} \)
(c) \( 2^{10} - 2 \)
(d) 10!
Answer: (b) \( 2^{10} \)
To form a 10-digit number using only the digits 2 and 3, each of the 10 places in the number can be filled in two ways (either with a 2 or a 3). Since there are 10 such places, we multiply the number of choices for each place. This is a basic principle of counting where choices are independent for each position.
Number of choices for the 1st digit = 2 (either 2 or 3)
Number of choices for the 2nd digit = 2 (either 2 or 3)
...
Number of choices for the 10th digit = 2 (either 2 or 3)
So, the total number of 10-digit numbers is \( 2 \times 2 \times 2 \times \ldots \) (10 times) \( = 2^{10} \).
In simple words: If you want to make a 10-digit number using only the digits 2 and 3, each spot in the number can be either a 2 or a 3. Since there are 10 spots, you multiply 2 by itself 10 times to get the total number of different numbers you can make.

🎯 Exam Tip: When forming numbers with a limited set of digits, consider the choices for each position. If repetition is allowed, multiply the number of choices for each position.

 

Question 22. If \( P_r \) stands for \( rP_r \), then the sum of the series \( 1 + P_1 + 2P_2 + 3P_3 + \ldots + nP_n \) is
(a) \( P_{n+1} \)
(b) \( P_{n+1} - 1 \)
(c) \( P_{n+1} + 1 \)
(d) \( (n+1)P_{n-1} \)
Answer: (a) \( P_{n+1} \)
Given that \( P_r = rP_r \), which means \( P_r = r! \).
So, the series is \( 1 + 1P_1 + 2P_2 + 3P_3 + \ldots + nP_n \).
Substitute \( P_r = r! \):
\( 1 + (1 \times 1!) + (2 \times 2!) + (3 \times 3!) + \ldots + (n \times n!) \)
We know the identity: \( r \times r! = (r+1)! - r! \). Let's use this for each term from \( 1P_1 \) onwards.
\( 1 + \sum_{r=1}^{n} (r \times r!) \)
\( = 1 + \sum_{r=1}^{n} ((r+1)! - r!) \)
This is a telescoping sum:
For \( r=1 \): \( (1+1)! - 1! = 2! - 1! \)
For \( r=2 \): \( (2+1)! - 2! = 3! - 2! \)
For \( r=3 \): \( (3+1)! - 3! = 4! - 3! \)
...
For \( r=n \): \( (n+1)! - n! \)
When we sum these up, all intermediate terms cancel out:
\( 1 + (2! - 1!) + (3! - 2!) + (4! - 3!) + \ldots + ((n+1)! - n!) \)
\( = 1 + (-1! + (n+1)!) \)
\( = 1 - 1 + (n+1)! \)
\( = (n+1)! \)
Since \( P_r = r! \), then \( P_{n+1} = (n+1)! \).
Therefore, the sum of the series is \( P_{n+1} \). This is a helpful trick to simplify factorial sums.
In simple words: The problem uses a special symbol \( P_r \) which is just \( r \) factorial. We need to add up a series where each term is \( r \) times \( r \) factorial. By using a clever math trick, we can see that all the middle parts cancel out. This leaves us with just \( (n+1) \) factorial, which is the same as \( P_{n+1} \).

🎯 Exam Tip: Recognizing telescoping sums, especially with factorials, can quickly simplify complex series problems. Remember the identity \( r \cdot r! = (r+1)! - r! \).

 

Question 23. The product of first n odd natural numbers equals
(a) \( 2nC_n \times nP_n \)
(b) \( (\frac{1}{2})^n 2nC_n \times nP_n \)
(c) \( (\frac{1}{4})^n \times 2nC_n \times 2nP_n \)
(d) \( nC_n \times nP_n \)
Answer: (b) \( (\frac{1}{2})^n 2nC_n \times nP_n \)
Let's find the product of the first n odd natural numbers. These are \( 1, 3, 5, \ldots, (2n-1) \).
The product \( = 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1) \).
To express this in terms of combinations and permutations, we can multiply and divide by the even numbers to make it a full factorial.
Multiply by \( (2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2n) \):
\( = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot (2n-1) \cdot (2n)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (2n)} \)
The numerator is \( (2n)! \).
The denominator can be written as \( 2^n \times (1 \cdot 2 \cdot 3 \cdot \ldots \cdot n) \), which is \( 2^n \times n! \).
So, the product \( = \frac{(2n)!}{2^n \times n!} \)
We know that \( 2nC_n = \frac{(2n)!}{n!n!} \) and \( nP_n = n! \).
Let's rearrange our result to match the options:
\( = \frac{(2n)!}{n! \cdot n!} \times \frac{n!}{2^n} \)
\( = 2nC_n \times \frac{n!}{2^n} \)
\( = (\frac{1}{2})^n \times 2nC_n \times n! \)
Since \( n! = nP_n \), we get:
\( = (\frac{1}{2})^n \times 2nC_n \times nP_n \)
This form matches option (b). This mathematical manipulation allows us to express a sequence of odd numbers using combinatorial terms.
In simple words: To multiply the first 'n' odd numbers, we can use a trick. We write it as a big factorial number divided by all the even numbers. After simplifying, it looks like a combination formula multiplied by a specific fraction involving 'n'.

🎯 Exam Tip: When dealing with products of odd or even numbers, try to expand them into a complete factorial by multiplying and dividing by the missing numbers. This often reveals a connection to permutations or combinations.

 

Question 24. If \( nC_4, nC_5, nC_6 \) are in A. P. the value of n can be
(a) 14
(b) 11
(c) 9
(d) 5
Answer: (a) 14
If three terms a, b, c are in an Arithmetic Progression (A.P.), then \( 2b = a+c \).
So, for \( nC_4, nC_5, nC_6 \) to be in A.P., we must have:
\( 2 \times nC_5 = nC_4 + nC_6 \)
Now, let's write out the combination formulas:
\( 2 \times \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} \)
Divide all terms by \( n! \) (assuming \( n \ge 6 \)):
\( \frac{2}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!} \)
Expand factorials to find common terms:
\( \frac{2}{5 \times 4!(n-5)!} = \frac{1}{4!(n-4)(n-5)!} + \frac{1}{6 \times 5 \times 4!(n-6)!} \)
Multiply by \( 4!(n-5)! \) (common denominator-related factor):
\( \frac{2}{5} = \frac{1}{n-4} + \frac{1}{6 \times 5} \times \frac{(n-5)!}{(n-6)!} \)
\( \frac{2}{5} = \frac{1}{n-4} + \frac{1}{30} \times (n-5) \)
Now multiply by 30(n-4) to clear denominators:
\( 12(n-4) = 30 + (n-5)(n-4) \)
\( 12n - 48 = 30 + n^2 - 4n - 5n + 20 \)
\( 12n - 48 = 30 + n^2 - 9n + 20 \)
\( 12n - 48 = n^2 - 9n + 50 \)
Rearrange into a quadratic equation:
\( n^2 - 9n - 12n + 50 + 48 = 0 \)
\( n^2 - 21n + 98 = 0 \)
Factor the quadratic equation:
\( n^2 - 14n - 7n + 98 = 0 \)
\( n(n-14) - 7(n-14) = 0 \)
\( (n-7)(n-14) = 0 \)
This gives two possible values for n: \( n = 7 \) or \( n = 14 \).
Both values are valid since \( n \ge 6 \) for the combinations to be defined. However, the options provided only list 14 as an answer. This is a common type of problem in combinatorics.
In simple words: When three combination numbers are in a pattern called an Arithmetic Progression, the middle one, multiplied by two, is equal to the sum of the other two. We use this rule and the combination formula. After some algebra, we get a simple equation that helps us find the possible values for 'n'. In this case, 'n' can be 7 or 14, and 14 is one of the options.

🎯 Exam Tip: Remember the property of arithmetic progressions: if a, b, c are in AP, then \( 2b = a+c \). When working with combinations, simplify by canceling common factorials, and be careful with algebraic manipulation.

 

Question 25. \( 1 + 3 + 5 + 7 + \ldots + 17 \) is equal to
(a) 101
(b) 81
(c) 71
(d) 61
Answer: (b) 81
The given series is an Arithmetic Progression (A.P.) because the difference between consecutive terms is constant.
First term \( a = 1 \).
Common difference \( d = 3 - 1 = 2 \).
Last term \( t_n = 17 \).
First, we need to find the number of terms (n) in this A.P. The formula for the nth term of an A.P. is \( t_n = a + (n-1)d \).
\( 17 = 1 + (n-1)2 \)
\( 16 = (n-1)2 \)
\( 8 = n-1 \)
\( \implies n = 9 \). So, there are 9 terms in the series.
Next, we find the sum of these terms using the formula for the sum of an A.P.: \( S_n = \frac{n}{2}[a + t_n] \).
\( S_9 = \frac{9}{2}[1 + 17] \)
\( S_9 = \frac{9}{2}[18] \)
\( S_9 = 9 \times 9 \)
\( S_9 = 81 \).
The sum of the first n odd natural numbers is simply \( n^2 \). Here, since there are 9 terms, the sum is \( 9^2 = 81 \).
In simple words: We have a list of odd numbers that start from 1 and go up to 17. To find their total sum, first, we count how many numbers are in the list. Then, we use a simple rule for adding up odd numbers, which is to square the count of numbers. There are 9 numbers in this list, and 9 times 9 is 81.

🎯 Exam Tip: For a series of consecutive odd numbers starting from 1, the sum is always equal to the square of the number of terms (\(n^2\)). This can save time compared to using the full A.P. sum formula.