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Sagot :
To solve this problem, we need to follow a step-by-step method based on the given sets and their operations. Let's calculate the necessary values one by one:
1. Given Sets and Operations:
- [tex]\( A \cup B = \{e, f, g, h, i\} \)[/tex]
- [tex]\( A - B = \{h, i\} \)[/tex]
- [tex]\( A \cap B = \{g\} \)[/tex]
2. Determine Set [tex]\( B \)[/tex]:
- We know that [tex]\( A \cup B \)[/tex] consists of elements in either [tex]\( A \)[/tex] or [tex]\( B \)[/tex].
- We also know [tex]\( A - B \)[/tex] gives elements in [tex]\( A \)[/tex] but not in [tex]\( B \)[/tex].
- [tex]\( A \cap B \)[/tex] provides the elements common to both [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
Since [tex]\( A \cup B = \{e, f, g, h, i\} \)[/tex] and [tex]\( A - B = \{h, i\} \)[/tex], it implies:
- Set [tex]\( B \)[/tex] must be made up of [tex]\( \{e, f, g\} \)[/tex] because [tex]\( h \)[/tex] and [tex]\( i \)[/tex] are not in [tex]\( B \)[/tex] and [tex]\( g \)[/tex] is common between [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
3. Calculate [tex]\( B - A \)[/tex]:
- [tex]\( B \)[/tex] includes [tex]\( \{e, f, g\} \)[/tex] if [tex]\( A \cap B = \{g\} \)[/tex], then [tex]\( B - A \)[/tex] (elements in [tex]\( B \)[/tex] but not in [tex]\( A \)[/tex]) must be [tex]\( \{e, f\} \)[/tex], as [tex]\( g \)[/tex] is removed as it is in both [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
4. Cardinality of [tex]\( B - A \)[/tex] [tex]\( (n(B - A)) \)[/tex]:
- The number of elements in [tex]\( B - A \)[/tex], which is [tex]\( \{e, f\} \)[/tex], is 2.
- So, [tex]\( n(B - A) = 2 \)[/tex].
5. Cardinality of [tex]\( B \)[/tex] [tex]\( (n(B)) \)[/tex]:
- The set [tex]\( B \)[/tex] includes [tex]\( \{e, f, g\} \)[/tex].
- So, [tex]\( n(B) = 3 \)[/tex].
6. Calculate [tex]\( n(B - A) + n(B) \)[/tex]:
- [tex]\( n(B - A) + n(B) = 2 + 3 = 5 \)[/tex].
Thus, the correct answer is:
[tex]\[ \boxed{5} \][/tex]
1. Given Sets and Operations:
- [tex]\( A \cup B = \{e, f, g, h, i\} \)[/tex]
- [tex]\( A - B = \{h, i\} \)[/tex]
- [tex]\( A \cap B = \{g\} \)[/tex]
2. Determine Set [tex]\( B \)[/tex]:
- We know that [tex]\( A \cup B \)[/tex] consists of elements in either [tex]\( A \)[/tex] or [tex]\( B \)[/tex].
- We also know [tex]\( A - B \)[/tex] gives elements in [tex]\( A \)[/tex] but not in [tex]\( B \)[/tex].
- [tex]\( A \cap B \)[/tex] provides the elements common to both [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
Since [tex]\( A \cup B = \{e, f, g, h, i\} \)[/tex] and [tex]\( A - B = \{h, i\} \)[/tex], it implies:
- Set [tex]\( B \)[/tex] must be made up of [tex]\( \{e, f, g\} \)[/tex] because [tex]\( h \)[/tex] and [tex]\( i \)[/tex] are not in [tex]\( B \)[/tex] and [tex]\( g \)[/tex] is common between [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
3. Calculate [tex]\( B - A \)[/tex]:
- [tex]\( B \)[/tex] includes [tex]\( \{e, f, g\} \)[/tex] if [tex]\( A \cap B = \{g\} \)[/tex], then [tex]\( B - A \)[/tex] (elements in [tex]\( B \)[/tex] but not in [tex]\( A \)[/tex]) must be [tex]\( \{e, f\} \)[/tex], as [tex]\( g \)[/tex] is removed as it is in both [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
4. Cardinality of [tex]\( B - A \)[/tex] [tex]\( (n(B - A)) \)[/tex]:
- The number of elements in [tex]\( B - A \)[/tex], which is [tex]\( \{e, f\} \)[/tex], is 2.
- So, [tex]\( n(B - A) = 2 \)[/tex].
5. Cardinality of [tex]\( B \)[/tex] [tex]\( (n(B)) \)[/tex]:
- The set [tex]\( B \)[/tex] includes [tex]\( \{e, f, g\} \)[/tex].
- So, [tex]\( n(B) = 3 \)[/tex].
6. Calculate [tex]\( n(B - A) + n(B) \)[/tex]:
- [tex]\( n(B - A) + n(B) = 2 + 3 = 5 \)[/tex].
Thus, the correct answer is:
[tex]\[ \boxed{5} \][/tex]
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