Sets Questions

Let A and B be two sets containing 2 elements and 4 elements, respectively. The number of subsets A\(\times\)B having 3 or more elements is

Solution : Given, n(A) = 2 and n(B) = 4 \(\therefore\) n(A\(\times\)B) = 8 The number of subsets of (A\(times\)B) having 3 or more elements = \(^8C_3 + {^8C_4} + ….. + {^8C_8}\) = \(2^8 – {^8C_0} – {^8C_1} – {^8C_2}\) = 256 – 1 – 8 – 28 = 219     [\(\because\) \(2^n\) = …

Let A and B be two sets containing 2 elements and 4 elements, respectively. The number of subsets A\(\times\)B having 3 or more elements is Read More »

Let A = [x: x \(\in\) R, |x| < 1]; B = [x : x \(\in\) R, |x - 1| \(\ge\) 1] and A \(\cup\) B = R - D, then the set D is

Solution : A = [x: x \(\in\) R,-1 < x < 1] B = [x : x \(\in\) R, x – 1 \(\le\) -1 or x – 1 \(\ge\) 1] [x: x \(\in\) R, x \(\le\) 0 or x \(\ge\) 2] \(\therefore\) A \(\cup\) B = R – D where D = [x : x …

Let A = [x: x \(\in\) R, |x| < 1]; B = [x : x \(\in\) R, |x - 1| \(\ge\) 1] and A \(\cup\) B = R - D, then the set D is Read More »

If N is a set of first 10 natural numbers and a relation R is defined as a + 2b = 10 where a, b \(\in\) N. find inverse of R.

Solution : R = {(2, 4), (4, 3), (6, 2), (8, 1)} \(R^{-1}\) = {(4, 2), (3, 4), (2, 6), (1, 8)} Similar Questions If A = {x,y}, then the power set of A is If aN = {ax : x \(\in\) N}, then the set 6N \(\cap\) 8N is equal to Let A = …

If N is a set of first 10 natural numbers and a relation R is defined as a + 2b = 10 where a, b \(\in\) N. find inverse of R. Read More »