A simple set Ques.
How many sets A exist for which { 1 , 2 } ⊆ A ⊆ { 1 , 2 , 3 , 4 , 5 } ?
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6 solutions
the set A may contain no other element than 1,2 (or) just one extra element[3,4,5] (or) two extra element[(3,4),(4,5),(3,5)] or all the three{3&4&5}.thus no. of possible ways=1+3+3+1=8
A should at least have 1 and 2 in its set. But it can also have other elements, one, two, or three more elements (namely 3, 4, or 5). So we can have {1, 2}, or {1, 2, _ } (with 3, 4, or 5 in the _ spot, which gives us 3 posibilities), {1, 2, _, _ } with any combination of 3 digits which gives us another 3 posibilities, and finally {1, 2, 3, 4, 5} itself.
Let's call the set of {1,2} ( A ),the set of {1,2,3,4,5} ( B ).We have:
A / B = 3 , 4 , 5
{3,4,5} have 3 numbers.The power set of {a,b,c,...,n) is 2 n u m b e r s i n t h e s e t so the power set of {3,4,5} is 2 3 = 8
Use Pascal's triangle.
1
1,1
1,2,1
1,3,3,1
1,4,6,4,1
This problem has a solution on the 4th line. The solution to 1,2 in A in 1,2,3,4,5,6 lies on the 5th line.
How does that work?
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He solved it correctly but failed to provide a good explanation. According to the given data, we can conclude that all the possible sets that A can be must satisfy the following conditions:
∙ 1 , 2 ∈ A ∙ 2 ≤ n ( A ) ≤ 5 ∙ 3 , 4 , 5 maybe ∈ A
So, we have two elements of A fixed and we are free to make a set A such that n ( A ) = 2 , 3 , 4 , 5 by taking 0 , 1 , 2 , 3 elements from the set { 3 , 4 , 5 } .
This can be done in ( 0 3 ) + ( 1 3 ) + ( 2 3 ) + ( 3 3 ) = 8 ways.
Since we always try to be as lazy as possible, some people avoid calculating the binomials themselves and simply sum the elements of the required row from Pascal's Triangle.
set A contain at least {1,2} and other element is 3,4,5. three element total 2 2 2 =8 subset .
The answer is: A is the union A = { 1 , 2 } ∪ C where C is any subset of the set { 3 , 4 , 5 } , but it is well-known that the powerset of a set with n elements has cardinality 2 n hence we get 2 3 = 8 possibilies.