เพิ่มเติมA ⊂ B แล้ว 1. A – B = ø 2.

เพิ่มเติม
A ⊂ B แล้ว 1. A – B = ø
2. A∩B = A
3. A∪B = B

การหาจำนวนเซตแบบประยุกต์
1) กำหนด n(A) = n , n(B) = m โดยที่ A ⊂ B
จำนวนเซต X ซึ่ง A ⊂ X ⊂ B = 2 n(B) – n(A) = 2 m – n เซต
2) กำหนด n(A) = n , n(B) = m โดยที่ A ⊂ B
จำนวนเซต X ซึ่ง A Ë X แต่ X ⊂ B = 2 n(B) – 2 n(B) – n(A) =2 m- 2 m – n เซต

การหาจำนวนสมาชิกของเซตจำกัด
1) 2 เซต
– n (A∪B) = n(A) + n(B) –n(A∩B)
– n [(A-B) ∪ (B-A)] = n(A) + n(B) –2[n(A∩B)]
2) 3 เซต
– n (A∪B∪C) = n(A) + n(B) +n(C)-n(A∩B) – n(A∩C) – n(B∩C) + n(A∩B∩C)

ผลลัพธ์ (อังกฤษ) 1:

MoreA ⊂ B, then B = A-ø 1. 2. A∩B = A 3. A∪B = BTo find the number of the application set1. (A) = n), given n, n (B) = A ⊂ B m, by the.A number of set B ⊂ ⊂ X, where X = n 2 n (B) – (A) = 0.2 m – n set.2. (A) = n), given n, n (B) = A ⊂ B m, by the.A number of the set X, where B = X, but X Ë ⊂ 2 n 2 n (B) – (B) – (A) n = n m-2 m-2 set.To find the number of the members of the set limit.2 set 1)– n (A∪B) = n(A) + n(B) –n(A∩B)– n [(A-B) ∪ (B-A)] = n(A) + n(B) –2[n(A∩B)] 3 set 2)– n (A∪B∪C) = n(A) + n(B) +n(C)-n(A∩B) – n(A∩C) – n(B∩C) + n(A∩B∩C)

ผลลัพธ์ (อังกฤษ) 2:

Additional
1. A ⊂ B then A - B = Ø
2. A∩B ​​= A
= B 3. A∪B a set of Application 1) to n (A) = n, n (B) = m by. A ⊂ B of the set X, A ⊂ X ⊂ B = 2 n (B) - n (A) = 2 m - n set 2) define n (A) = n, n (B) = m, where A ⊂ B. A Ë X number of sets X, but X ⊂ B = 2 n (B) - 2 n (B) - n (A) = 2 m- 2 m - n sets a finite number of members of one), two sets - n. (A∪B) = n (A) + n (B) -n (A∩B) - n [(AB) ∪ (BA)] = n (A) + n (B) -2 [n (A∩. B)] 2) Set 3 - n (A∪B∪C) = n (A) + n (B) + n (C) -n (A∩B) - n (A∩C) - n (B∩. C) + n (A∩B∩C).

ผลลัพธ์ (อังกฤษ) 3:

More
A ⊂ B and 1. A - B = ø
2. A ∩ B = A
3. A ∪ B = B

finding the set applied model
1) to determine n (A) = n n (, B) = m by. A ⊂ B
were set X which A ⊂ X ⊂ B = 2 n (B) - n (A) = 2 m - n set
2) to determine n (A) = n n (, B) = m without A ⊂ B
.The set X which A Ë X but X ⊂ B = 2 n (B) - 2 n (B) - n (A) = 2 m - 2 m - n set

finding of element of a set limit
1) 2 set
- n (A ∪ B). = n (A) n (B) - n (A ∩ B)
- n [(A-B) ∪ (B-A)] = n (A) n (B) - 2 [n (A ∩ B)]
2) 3 set
- n (A ∪ B ∪ C) = n (A) n (B) n (C) - n (A ∩ B) - n (A ∩ C). To n (B ∩ C) n (A ∩ B ∩ C)
.