Number of Element in Set

Number of elements in a set

The number of elements in set A is denoted by $n(A)$

Example

Let A be a set of all positive odd integers which are less than 10. Find $n(A)$

Solution

$A = \{1, 3, 5, 7, 9\}$

Now $n(A) = 5$

Example

Let $A = \{x \in \mathbb{R} : x^2 - x - 2 = 0\}$. Find $n(A)$

Solution

$$\begin{aligned} &x^2 - x - 2 = 0 \\ &a \times c = -2 \\ &x^2 - 1x - 2x - 2 = 0 \\ &x(x + 1) - 2(x + 1) = 0 \\ &(x - 2)(x + 1) = 0 \\ &x = 2, \ x = -1 \\ &A = \{-1, 2\} \\ &\text{now } n(A) = 2 \end{aligned}$$

Note

1. The number of elements of a set is defined only for a finite set
2. If $A \subseteq U$ then the number of elements of $A'$ is $n(A') = n(U) - n(A)$

Example

If $A \subseteq U$ and $B \subseteq U$ then show that $n(A \cup B) = n(A) + n(B) - n(A \cap B)$

Proof

Refer to the Venn diagram below

$$\begin{aligned} \text{R.H.S} &= n(A) + n(B) - n(A \cap B) \\ &= (a + b) + (b + c) - b \\ &= a + 2b + c - b \\ &= a + b + c \\ &= n(A \cup B) \\ &= \text{L.H.S} \end{aligned}$$

Example

1. Given $n(X) = 18$, $n(Y) = 26$, $n(X \cap Y) = 12$. Find $n(X \cup Y)$
2. Given $n(S \cup T) = 19$, $n(S) = 15$, $n(S \cap T') = 10$. Find $n(S \cap T)$
3. Given $n(A \cap B) = 5$, $n(A \cup B) = 16$, $n((A \cup B)') = 4$, $n(A - B) = 8$. Find i) $n(A)$ ii) $n(A \cup B')$ iii) $n(\mu)$ iv) $n(A' \cap B)$

Solutions

$$\begin{aligned} n(X \cup Y) &= n(X) + n(Y) - n(X \cap Y) \\ &= 18 + 26 - 12 \\ &= 32 \\ n(X \cup Y) &= 32 \end{aligned}$$

2. $n(S \cup T) = 19$, $n(S) = 15$, $n(S \cap T') = 10$

$n(S \cap T) = ?$

$$\begin{aligned} n(S \cap T) &= n(S) - n(S \cap T') \\ &= 15 - 10 \\ &= 5 \\ n(S \cap T) &= 5 \end{aligned}$$

3. $n(A \cap B) = 5$, $n(A \cup B) = 16$, $n((A \cup B)') = 4$, $n(A - B) = 8$

i) $n(A) = ?$ ii) $n(A \cup B')$ iii) $n(\mu)$ iv) $n(A' \cap B)$

$$\begin{aligned} n(A) &= n(A - B) + n(A \cap B) \\ &= 8 + 5 \\ &= 13 \end{aligned}$$

ii)

$$\begin{aligned} n(A \cup B') &= n(A) + n(B') \\ &= 13 + 4 \\ &= 17 \\ n(A \cup B') &= 17 \end{aligned}$$

iii)

$$\begin{aligned} n(\mu) &= n(A \cup B) + n((A \cup B)') \\ &= 16 + 4 \\ &= 20 \\ n(\mu) &= 20 \end{aligned}$$

iv)

$$\begin{aligned} n(A' \cap B) &= n(B) - n(A \cap B) \\ n(A' \cap B) &= n(A \cup B) - n(A) + n(A \cap B) - n(A \cap B) \\ n(A' \cap B) &= 16 - 13 + 5 - 5 \\ n(A' \cap B) &= 3 \end{aligned}$$

4. By using $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ show that:

$$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)$$

Solutions

Let $B \cup C = K$

L.H.S $n(A \cup B \cup C) = n(A \cup K)$

$$\begin{aligned} &= n(A) + n(K) - n(A \cap K) \\ &= n(A) + n(B \cup C) - n(A \cap (B \cup C)) \\ &= n(A) + n(B) + n(C) - n(B \cap C) - n((A \cap B) \cup (A \cap C)) \\ &= n(A) + n(B) + n(C) - n(B \cap C) - (n(A \cap B) + n(A \cap C) - n((A \cap B) \cap (A \cap C))) \\ &= n(A) + n(B) + n(C) - n(B \cap C) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C) \end{aligned}$$

Example

1. A class has 15 boys and 15 girls. In the class 20 students are studying science, 14 students are studying math, 10 boys are studying science, 10 boys are studying math, 8 boys are studying both math and science, 4 girls are studying neither math nor science.

Find

i) How many students study math only?

ii) How many students study science only?

iii) How many students study both math and science?

2. In a class of 35 students each student takes either one of two subjects (physics, chemistry and biology). If 13 students take chemistry, 22 students take physics, 17 students take biology, 6 students take both physics and chemistry and 3 students take both biology and chemistry. Find the number of students who take both biology and physics.

Solution for problem 1

i) Students who study math only = $2 + 1$ = 3 students

ii) Students who study science only = $2 + 7$ = 9 students

iii) Students who study both math and science = $8 + 3$ = 11 students

Solution for problem 2

$18 - x = 15$

$x = 3$