The power set is the set of all subsets of A. P(A) = { {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} }. The number of elements n(P(A)) = 2^n = 2^3 = 8.

n(P(A)) = 2^n

Let n(C) = 24, n(F) = 16, n(C union F) = 35 (each likes at least one). By the formula n(C union F) = n(C) + n(F) - n(C intersection F), we get 35 = 24 + 16 - n(C intersection F), so n(C intersection F) = 40 - 35 = 5. Hence 5 students like both games.

n(A union B) = n(A) + n(B) - n(A intersection B)

Union (A union B) is the set of all elements that are in A or in B or in both. Intersection (A intersection B) is the set of elements common to both A and B. Difference (A - B) is the set of elements that are in A but not in B. Example: if A = {1,2,3} and B = {2,3,4}, then A union B = {1,2,3,4}, A intersection B = {2,3}, and A - B = {1}.

{x : -3 <= x < 5} includes -3 but not 5, so in interval notation it is [-3, 5). The interval (2, 9] excludes 2 and includes 9, so in set-builder form it is {x : x is a real number, 2 < x <= 9}.

The Cartesian product A x B is the set of all ordered pairs (a, b) with a in A and b in B. A x B = {(1,3), (1,4), (2,3), (2,4)}. The number of elements n(A x B) = n(A) x n(B) = 2 x 2 = 4.

n(A x B) = n(A) x n(B)

Two ordered pairs are equal when their corresponding components are equal. So x + 1 = 3 gives x = 2, and y - 2 = 1 gives y = 3. Hence x = 2 and y = 3.

(a, b) = (c, d) => a = c and b = d

The domain of a relation is the set of all first components (inputs), and the range is the set of all second components (outputs). For R = {(1,2), (2,4), (3,6)}, the domain = {1, 2, 3} and the range = {2, 4, 6}.

A function f from A to B is one-one (injective) if different elements of A have different images in B, i.e. f(x1) = f(x2) implies x1 = x2. A function f from A to B is onto (surjective) if every element of B is the image of at least one element of A, i.e. the range of f equals the codomain B.

Using pi radians = 180 degrees: 5pi/6 radians = (5/6) x 180 = 150 degrees. And 240 degrees = 240 x (pi/180) = 4pi/3 radians.

radians = degrees x (pi/180)

Write 75 = 45 + 30. Using sin(A + B) = sin A cos B + cos A sin B: sin 75 = sin 45 cos 30 + cos 45 sin 30 = (1/sqrt2)(sqrt3/2) + (1/sqrt2)(1/2) = (sqrt3 + 1)/(2 sqrt2) = (sqrt6 + sqrt2)/4.

sin(A + B) = sin A cos B + cos A sin B

Start from the fundamental identity sin^2 x + cos^2 x = 1. Divide both sides by cos^2 x (cos x not equal to 0): sin^2 x/cos^2 x + cos^2 x/cos^2 x = 1/cos^2 x, which gives tan^2 x + 1 = sec^2 x. Hence 1 + tan^2 x = sec^2 x.

1 + tan^2 x = sec^2 x

Write 15 = 45 - 30. Using sin(A - B) = sin A cos B - cos A sin B: sin 15 = sin 45 cos 30 - cos 45 sin 30 = (1/sqrt2)(sqrt3/2) - (1/sqrt2)(1/2) = (sqrt3 - 1)/(2 sqrt2) = (sqrt6 - sqrt2)/4.

sin(A - B) = sin A cos B - cos A sin B

(1 + i)(2 - i) = 1 x 2 + 1 x (-i) + i x 2 + i x (-i) = 2 - i + 2i - i^2. Since i^2 = -1, this is 2 + i + 1 = 3 + i. So a = 3 and b = 1.

i^2 = -1

The multiplicative inverse of z is 1/z = conjugate(z)/|z|^2. Here conjugate of (4 - 3i) is (4 + 3i) and |z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25. So the inverse = (4 + 3i)/25 = 4/25 + (3/25) i.

z^-1 = conjugate(z)/|z|^2

Using the quadratic formula x = [-b +/- sqrt(b^2 - 4ac)]/(2a) with a = 1, b = 1, c = 1: discriminant = 1 - 4 = -3. So x = [-1 +/- sqrt(-3)]/2 = [-1 +/- i sqrt3]/2. The roots are (-1 + i sqrt3)/2 and (-1 - i sqrt3)/2.

x = [-b +/- sqrt(b^2 - 4ac)]/(2a)

Powers of i repeat with period 4 (i, -1, -i, 1). i^9 = i^(8+1) = (i^4)^2 x i = 1 x i = i. i^19 = i^(16+3) = (i^4)^4 x i^3 = 1 x (-i) = -i. So i^9 + i^19 = i + (-i) = 0.

i^4 = 1

3x - 5 < 7 gives 3x < 12, so x < 4. The solution set is {x : x < 4, x is real} = (-infinity, 4). On the number line, this is shown by an open circle at 4 with shading to the left (all values less than 4).

Write 8! = 8 x 7 x 6!. So 8!/(6! x 2!) = (8 x 7 x 6!)/(6! x 2!) = (8 x 7)/2! = 56/2 = 28.

n! = n x (n-1)!

The hundreds place can be filled in 9 ways (any of 1 to 9), the tens place in 8 ways (remaining digits) and the units place in 7 ways. By the multiplication principle, total numbers = 9 x 8 x 7 = 504.

total = product of choices at each place

Since order does not matter in a committee, we use combinations. Number of ways = 10C3 = 10!/(3! x 7!) = (10 x 9 x 8)/(3 x 2 x 1) = 720/6 = 120.

nCr = n!/(r!(n-r)!)