The problem gives the number of elements in the universal set $\xi$, $X$, and $Y$, denoted as $n(\xi) = 20$, $n(X) = 15$, and $n(Y) = 8$. We need to find: (a) $n(Y')$, the number of elements in the complement of $Y$. (b) the largest possible value of $n(X \cap Y)$, the number of elements in the intersection of $X$ and $Y$. (c) the smallest possible value of $n(X \cup Y)$, the number of elements in the union of $X$ and $Y$.

1. Problem Description

The problem gives the number of elements in the universal set $\xi$, $X$, and $Y$, denoted as $n(\xi) = 20$, $n(X) = 15$, and $n(Y) = 8$. We need to find:

(a) $n(Y')$, the number of elements in the complement of $Y$.

(b) the largest possible value of $n(X \cap Y)$, the number of elements in the intersection of $X$ and $Y$.

(c) the smallest possible value of $n(X \cup Y)$, the number of elements in the union of $X$ and $Y$.

2. Solution Steps

(a) To find the number of elements in the complement of $Y$, we use the formula:

$n(Y') = n(\xi) - n(Y)$

$n(Y') = 20 - 8 = 12$

(b) The largest possible value of $n(X \cap Y)$ occurs when $Y$ is a subset of $X$. In this case, $n(X \cap Y)$ will be equal to $n(Y)$.

$n(X \cap Y)_{max} = \min(n(X), n(Y))$

$n(X \cap Y)_{max} = \min(15, 8) = 8$

(c) The smallest possible value of $n(X \cup Y)$ occurs when $X$ and $Y$ have the largest possible intersection. We use the formula:

$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$

To minimize $n(X \cup Y)$, we need to maximize $n(X \cap Y)$. We already know that the largest possible value of $n(X \cap Y)$ is

8. So,

$n(X \cup Y)_{min} = n(X) + n(Y) - n(X \cap Y)_{max}$

$n(X \cup Y)_{min} = 15 + 8 - 8 = 15$

Also, we need to ensure that $n(X \cup Y) \le n(\xi)$. Since 15 is less than or equal to 20, this value is valid.

3. Final Answer

(a) 12

(b) 8

(c) 15