## Permutation and Combination – Class XI – Exercise 7.4

If nC8 = nC2. find nC2 Solution: We know that, nCa = nCb then a = b or n = a + b Therefore, nC8 = nC2, since 8≠2 , we have 8 + 2 = 10 nC2 = 10C2 = 10!/2!(10-2)! = 10!/2!8! = 10x9x8/2x1x8 = 45 Determine n if (i) (2nC3 :… Continue reading Permutation and Combination – Class XI – Exercise 7.4

## Complex Numbers and Quadratic Equations – Class XI – Exercise 5.2

1.Find the modulus and the argument of the complex number z = -1 – i √3 Solution: z = -1 - i√3 Let r cosθ = -1 and r sinθ = -√3 On squaring and adding, we get (r cosθ)2 + ( r sinθ)2 = (-1)2 + (-√3)2 r2(cos2θ + sin2θ) = 1 +… Continue reading Complex Numbers and Quadratic Equations – Class XI – Exercise 5.2

## Linear Inequalities – Class XI – Exercise 6.1 [1 0 – 26]

Solve the given inequality for real x: 3(x-2)/5 ≤ 5(2-x)/3 Solution: The given inequality is 3(x-2)/5 ≤ 5(2-x)/3 ⇒ 3(x-2)/5 ≤ 5(2-x)/3 ⇒ 9(x-2) ≤25(2-x) ⇒ 9x – 18 ≤ 50 – 25x ⇒ 34x – 18 ≤ 50 ⇒ 34x ≤ 68 ⇒ 34x/34 ≤ 68/34 ⇒ x ≤ 2 Thus, all real numbers… Continue reading Linear Inequalities – Class XI – Exercise 6.1 [1 0 – 26]

## Permutation and Combination – Class XI – Exercise 7.3

How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? Solution: 3-digit numbers have to be formed using the digits 1 to 9. Here, the order of the digits matters. Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits… Continue reading Permutation and Combination – Class XI – Exercise 7.3

## Permutation and Combination – Class XI – Exercise 7.1

1: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) Repetition of the digits is allowed? (ii) Repetition of the digits is not allowed? Solution: (i) There will be number of ways of filling 3 vacant places in succession by the given five digits. Here,… Continue reading Permutation and Combination – Class XI – Exercise 7.1