Samacheer Kalvi Class 11 Business Maths Solutions Chapter 2 Algebra Exercise 2.7

Question 1. If \( nC_3 = nC_2 \) then the value of \( nC_4 \) is:
(a) 3
(b) 4
(c) 5
(d) 5
Answer: (d) 5
In simple words: If two combinations \( nC_x \) and \( nC_y \) are equal, and \( x \) is not equal to \( y \), then \( n \) must be the sum of \( x \) and \( y \). So, \( n = 3 + 2 = 5 \). Then we need to find \( 5C_4 \), which is 5.

🎯 Exam Tip: Remember the key property of combinations: if \( nC_x = nC_y \) and \( x \neq y \), then \( x+y=n \). This saves time in finding 'n'.

Question 2. The value of n, when \( nP_2 = 20 \) is:
(a) 3
(b) 6
(c) 5
(d) 4
Answer: (c) 5
In simple words: We are looking for a number \( n \) such that when you multiply it by the number just before it ( \( n-1 \) ), you get 20. The numbers 5 and 4 multiply to 20, so \( n \) must be 5.

🎯 Exam Tip: For small permutation values like \( nP_2 \), directly expressing it as \( n(n-1) \) and then trying small integer values for 'n' can be faster than quadratic equation solving.

Question 3. The number of ways selecting 4 players out of 5 is:
(a) 20
(b) 25
(c) 5
Answer: (c) 5
In simple words: To choose 4 players from a group of 5, it's the same as choosing which 1 player you will NOT pick. Since there are 5 players in total, there are 5 ways to choose the one to leave out, which means there are 5 ways to pick 4 players.

🎯 Exam Tip: Remember the combination identity \( nC_r = nC_{n-r} \). It simplifies calculations, for example, \( 5C_4 \) is much easier to compute as \( 5C_1 \).

Question 4. If \( nP_r = 720(nC_r) \), then r is equal to:
(a) 4
(b) 5
(c) 6
(d) 7
Answer: (c) 6
In simple words: Permutations are how many ways you can arrange things, and combinations are how many ways you can pick them. The formula \( nP_r = r! \times nC_r \) tells us the connection. In this problem, we are given \( nP_r = 720 \times nC_r \). If we compare this to the formula, we see that \( r! \) must be 720. We then find that 6 multiplied by all numbers before it down to 1 (which is 6 factorial) equals 720. So, \( r \) is 6.

🎯 Exam Tip: Remember the relationship \( nP_r = r! \times nC_r \). It is a direct way to solve problems involving both permutations and combinations.

Question 5. The possible outcomes when a coin is tossed five times:
(a) 25
(b) 10
(c) \( \frac{5}{2} \)
Answer: (a) 25
In simple words: A coin has 2 sides. If you flip it 5 times, there are 25 different ways it can land, according to the options given.

🎯 Exam Tip: For independent events, the total number of outcomes is found by multiplying the number of outcomes for each individual event.

Question 6. The number of diagonals in a polygon of n sides is equal to:
(a) \( nC_2 \)
(b) \( nC_2 - 2 \)
(c) \( nC_2 - n \)
(d) \( nC_2 - 1 \)
Answer: (c) \( nC_2 - n \)
In simple words: For a shape with \( n \) corners, you can draw a line between any two corners in \( nC_2 \) ways. Some of these lines are the sides of the shape itself. Since there are \( n \) sides, you remove these \( n \) lines, and what's left are the diagonals.

🎯 Exam Tip: Remember that the total combinations of any two vertices in an \( n \)-sided polygon includes both the sides and the diagonals. Subtracting the \( n \) sides gives the number of diagonals.

Question 7. The greatest positive integer which divide \( n(n + 1) (n + 2) (n + 3) \) for all \( n \in N \) is:
(a) 2
(b) 6
(c) 20
(d) 24
Answer: (d) 24
In simple words: We are looking for the biggest number that can always divide the product of any four numbers that come one after another. If you try with \( n=1 \), the numbers are 1, 2, 3, and 4. Their product is 24. Any product of four consecutive integers will always have factors of 1, 2, 3, and 4, so it must be divisible by 24.

🎯 Exam Tip: To find the greatest common divisor for an expression involving consecutive integers, substitute the smallest possible integer (usually 1 or 0) and calculate the result. This will often give the answer related to factorials.

Question 8. If n is a positive integer, then the number of terms in the expansion of \( (x + a)^n \) is:
(a) n
(b) n + 1
(c) n - 1
(d) 2n
Answer: (b) n + 1
In simple words: When you expand something like \( (x+a) \) raised to the power of \( n \), you will always get \( n+1 \) terms in the answer. For example, if the power is 2, you get 3 terms.

🎯 Exam Tip: A common mistake is to think the number of terms is equal to 'n'. Remember to always add 1 to the power 'n' to get the total number of terms.

Question 9. For all n > 0, \( nC_1 + nC_2 + nC_3 + ...... + nC_n \) is equal to:
(a) \( 2n \)
(b) \( 2^n - 1 \)
(c) \( n^2 \)
(d) \( n^2 - 1 \)
Answer: (b) \( 2^n - 1 \)
In simple words: The sum of all binomial coefficients from \( nC_0 \) to \( nC_n \) is always \( 2 \) raised to the power of \( n \). If we only sum from \( nC_1 \) (meaning we don't include \( nC_0 \) ), then we subtract the value of \( nC_0 \), which is 1. So, the sum becomes \( 2^n - 1 \).

🎯 Exam Tip: Remember the identity \( \sum_{r=0}^{n} nC_r = 2^n \). If the sum starts from \( nC_1 \), always subtract \( nC_0 \) (which is 1) from \( 2^n \).

Question 10. The term containing \( x^3 \) in the expansion of \( (x – 2y)^7 \) is:
(a) 3rd
(b) 4th
(c) 5th
Answer: (c) 5th
In simple words: When you expand \( (x - 2y)^7 \), each term has powers of \( x \) and \( y \). The total power for \( x \) and \( y \) always adds up to 7. If we want \( x^3 \), then the power of \( y \) must be \( 7-3=4 \). The term number is always one more than the power of the second part (here, \( -2y \) ). So, if the power of \( -2y \) is 4, the term is the 5th term.

🎯 Exam Tip: Use the general term formula \( T_{r+1} = nC_r x^{n-r} a^r \). Set the exponent of 'x' or 'a' to the desired value and solve for 'r', then the term number is \( r+1 \).

Question 11. The middle term in the expansion of \( \left(x+\frac{1}{x}\right)^{10} \) is:
(a) \( 10C_4 \left(\frac{1}{x}\right) \)
(b) \( 10C_5 \)
(c) \( 10C_6 \)
(d) \( 10C_7 x^2 \)
Answer: (b) \( 10C_5 \)
In simple words: When the power \( n \) is an even number, there is one middle term. Its position is found by taking half of \( n \) and adding 1. For \( n=10 \), the middle term is the 6th term. When you calculate the 6th term for \( (x + \frac{1}{x})^{10} \), the \( x^5 \) and \( \frac{1}{x^5} \) parts multiply to 1, leaving only the combination part, \( 10C_5 \).

🎯 Exam Tip: For even powers 'n', the middle term is always \( T_{\frac{n}{2}+1} \). For expressions like \( (x + \frac{1}{x})^n \), the powers of \( x \) often cancel out in the middle term.

Question 12. The constant term in the expansion of \( \left(x+\frac{2}{x}\right)^{6} \) is:
(a) 156
(b) 165
(c) 162
(d) 160
Answer: (d) 160
In simple words: We need to find the term in the expansion of \( (x + \frac{2}{x})^6 \) that has no \( x \) in it. Since the power is 6 (an even number), the middle term will be the one without \( x \). This is the 4th term. When we calculate the 4th term, the \( x^3 \) in the numerator and denominator cancel out. The remaining numbers multiply to 160.

🎯 Exam Tip: For binomials of the form \( \left(ax + \frac{b}{x}\right)^n \), the constant term will always be a middle term if 'n' is even. If 'n' is odd, there is no constant term of this form unless 'x' has different powers like \( x^p \) and \( 1/x^q \).

Question 13. The last term in the expansion of \( (3 + \sqrt{2})^8 \) is:
(a) 81
(b) 16
(c) 8
(d) 2
Answer: (b) 16
In simple words: In any expansion like \( (A+B)^n \), the very last term is just \( B \) raised to the power of \( n \). Here, \( B \) is \( \sqrt{2} \) and \( n \) is 8. So, the last term is \( (\sqrt{2})^8 \). This means multiplying \( \sqrt{2} \) by itself 8 times, which simplifies to \( 2^4 \), or 16.

🎯 Exam Tip: For a binomial \( (A+B)^n \), the first term is \( A^n \) and the last term is \( B^n \). Don't confuse the terms' positions with their values.

Question 14. If \( \frac{k x}{(x+4)(2 x-1)}=\frac{4}{x+4}+\frac{1}{2 x-1} \) then k is equal to:
(a) 9
(b) 11
(c) 7
Answer: (a) 9
In simple words: We have an equation where fractions are equal. We can combine the two fractions on the right side into one. To do this, we multiply the top of each fraction by the bottom of the other. After combining and simplifying, we get \( \frac{9x}{(x+4)(2x-1)} \). Comparing this to the left side, \( \frac{kx}{(x+4)(2x-1)} \), we can see that \( k \) must be 9.

🎯 Exam Tip: When dealing with equations involving fractions, a common strategy is to make the denominators on both sides equal, allowing you to equate the numerators to solve for unknown variables.

Question 15. The number of 3 letter words that can be formed from the letters of the word ‘NUMBER' when the repetition is allowed are:
(a) 206
(b) 133
(c) 216
(d) 300
Answer: (c) 216
In simple words: The word 'NUMBER' has 6 different letters. We want to make a 3-letter word, and we can use the same letter more than once. For the first letter, we have 6 choices. For the second letter, we still have 6 choices because we can repeat. For the third letter, we also have 6 choices. So, we multiply \( 6 \times 6 \times 6 \), which gives us 216 different 3-letter words.

🎯 Exam Tip: When repetition is allowed, the number of choices for each position remains constant. If 'n' items are chosen 'r' times with repetition, the total ways are \( n^r \). Be careful to count the *distinct* letters if the word has repeated letters.

Question 16. The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is:
(a) 18
(b) 12
(c) 9
Answer: (a) 18
In simple words: To make a parallelogram, you need two parallel lines running one way, and two other parallel lines running a different way that cross the first two. We have 4 lines in the first set, so we pick 2 of them (\( 4C_2 \) ways). We have 3 lines in the second set, so we pick 2 of those (\( 3C_2 \) ways). Then we multiply the number of ways from each set to get the total number of parallelograms, which is \( 6 \times 3 = 18 \).

🎯 Exam Tip: The number of parallelograms formed by 'm' parallel lines and 'n' parallel lines intersecting them is always \( mC_2 \times nC_2 \).

Question 17. There are 10 true or false questions in an examination. Then these questions can be answered in
(a) 240 ways
(b) 120 ways
(c) 1024 ways
(d) 100 ways
Answer: (c) 1024 ways
In simple words: Each of the 10 questions can be answered in 2 ways (true or false). To find the total ways to answer all 10 questions, we multiply 2 by itself 10 times. This means \( 2^ {10} \), which equals 1024.

🎯 Exam Tip: For 'n' true/false questions, or any scenario with 'n' independent choices each having 2 options, the total number of ways is \( 2^n \).

Question 18. The value of \( (5C_0 + 5C_1) + (5C_1 + 5C_2) + (5C_2 + 5C_3) + (5C_3 + 5C_4) + (5C_4 + 5C_5) \) is:
(a) \( 2^6 - 2 \)
(b) \( 2^5 - 1 \)
(c) \( 2^8 \)
(d) \( 2^7 \)
Answer: (a) \( 2^6 - 2 \)
In simple words: We are adding pairs of combination numbers. If we group them, we have the full sum of \( 5C_r \) (which is \( 2^5 \)) plus almost another full sum, but missing the first and last parts. The sum of all \( 5C_r \) is \( 2^5 \). The second part is \( 2^5 - 5C_0 - 5C_5 \). Since \( 5C_0 \) and \( 5C_5 \) are both 1, the second part is \( 2^5 - 2 \). So, the total sum is \( 2^5 + (2^5 - 2) \), which simplifies to \( 2 \times 2^5 - 2 = 2^6 - 2 \).

🎯 Exam Tip: Break down complex sums of binomial coefficients into standard known identities like \( \sum_{r=0}^{n} nC_r = 2^n \). Look for terms that appear multiple times or terms that are missing from a full sum.

Question 19. The total number of 9 digit number which has all different digit is:
(a) 10!
(b) 9!
(c) \( 9 \times 9! \)
(d) \( 10 \times 10! \)
Answer: (c) \( 9 \times 9! \)
In simple words: We want to make a 9-digit number using 10 different digits (0 to 9), without repeating any digit. The first digit can't be zero, so there are 9 options for it. For the second digit, we have 9 options left (the one we didn't use, plus zero). For the third, there are 8 options, and so on. This means we multiply 9 (for the first digit) by the permutation of the remaining 9 digits, which is \( 9! \). So the total is \( 9 \times 9! \).

🎯 Exam Tip: When forming numbers with distinct digits, always consider the restriction that the first digit cannot be zero. Calculate the choices for the first position separately, then handle the remaining positions as a permutation of the remaining available digits.

Question 20. The number of ways to arrange the letters of the word “CHEESE":
(a) 120
(b) 240
(c) 720
(d) 6
Answer: (a) 120
In simple words: The word "CHEESE" has 6 letters. But the letter 'E' appears 3 times. To find the number of different ways to arrange these letters, we calculate the factorial of the total number of letters (6!), and then divide it by the factorial of the number of times each repeated letter appears (3! for 'E'). So, it's \( 6! \div 3! \), which equals 120.

🎯 Exam Tip: When letters in a word repeat, remember to divide the total factorial of the letters by the factorial of the count of each repeating letter.

Question 21. 13 persons participated in a dinner. The number of handshakes that happened in the dinner is:
(a) 715
(b) 78
(c) 286
(d) 13
Answer: (b) 78
In simple words: When people shake hands, two people are involved in each handshake. If there are 13 people, we need to find how many ways we can choose any 2 people from this group. This is a combination problem \( 13C_2 \). You multiply 13 by 12, then divide by 2, which gives 78 handshakes.

🎯 Exam Tip: The number of handshakes among 'n' people is always \( nC_2 \), because each handshake involves exactly two people and the order of selection doesn't matter.

Question 22. The number of words with or without meaning that can be formed using letters of the word "EQUATION”, with no repetition of letters is:
(a) 7!
(b) 3!
(c) 8!
(d) 5!
Answer: (c) 8!
In simple words: The word "EQUATION" has 8 different letters. If we want to arrange all 8 letters to make new words, and we can't repeat any letter, the number of ways is 8 factorial (8!). This means multiplying 8 by every whole number down to 1.

🎯 Exam Tip: If you have 'n' distinct items and you need to arrange all 'n' of them, the number of ways is simply \( n! \).

Question 23. Sum of binomial coefficient in a particular expansion is 256, then number of terms in the expansion is:
(a) 8
(b) 7
(c) 6
(d) 9
Answer: (a) 8
In simple words: The total sum of all the numbers in a binomial expansion is \( 2 \) raised to the power of \( n \). If this sum is 256, then \( 2^n = 256 \), which means \( n \) must be 8. The problem asks for this value of \( n \).

🎯 Exam Tip: Be careful to distinguish between the power of the expansion 'n' and the total number of terms, which is always 'n+1'. In some cases, questions may implicitly ask for 'n'.

Question 24. The number of permutation of n different things taken r at a time, when the repetition is allowed is:
(a) \( r^n \)
(b) \( n^r \)
(c) \( \frac{n!}{(n-r)!} \)
(d) \( \frac{n!}{(n+r)!} \)
Answer: (b) \( n^r \)
In simple words: If you have \( n \) different things and you want to pick \( r \) of them, one after another, and you can use the same thing multiple times, then for each of the \( r \) picks, you have \( n \) choices. So, you multiply \( n \) by itself \( r \) times, which is \( n^r \).

🎯 Exam Tip: Differentiate between permutations with and without repetition. With repetition allowed, it's \( n^r \); without repetition, it's \( nP_r = \frac{n!}{(n-r)!} \).

Question 25. The sum of the binomial coefficients is:
(a) \( 2n \)
(b) \( n^2 \)
(c) \( 2^n \)
(d) \( n + 17 \)
Answer: (c) \( 2^n \)
In simple words: If you add up all the binomial coefficients for any power \( n \) (starting from \( nC_0 \) all the way to \( nC_n \) ), the total sum will always be equal to \( 2 \) raised to the power of \( n \).

🎯 Exam Tip: This identity, \( \sum_{r=0}^{n} nC_r = 2^n \), is crucial and frequently used in combinatorics and probability. Ensure you remember it.