Let S be a set with 10 elements. The number of subsets of S having odd...
Number of subset having odd number of elements
=
10C
1 +
10C
2 + ... +
10C
9 =
= 512
Since Z
I2 under addition modulo 12 is a cyclic group
So, Total number of non-trivial proper subgroups
= Z(12) - 2
= 6 - 2 = 4
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Let S be a set with 10 elements. The number of subsets of S having odd...
Explanation:
To find the number of subsets of S having an odd number of elements, we can use the concept of the power set.
Power Set: The power set of a set S is the set of all possible subsets of S, including the empty set and the set itself. The power set of a set with n elements has 2^n subsets.
In this case, since the set S has 10 elements, the power set of S will have 2^10 = 1024 subsets in total.
Now, let's consider the subsets with an odd number of elements. We can observe that if a subset has an odd number of elements, then its complement (the set of elements not included in the subset) will also have an odd number of elements.
For example, if a subset has 1 element, its complement will have 9 elements. If a subset has 3 elements, its complement will have 7 elements. And so on.
Since the power set contains all possible subsets, for every subset with an odd number of elements, its complement will also be present in the power set.
Therefore, half of the subsets in the power set will have an odd number of elements.
So, the number of subsets of S having an odd number of elements is 1/2 * (total number of subsets) = 1/2 * 1024 = 512.
Therefore, the correct answer is option B) 512.