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Updated on: 19 Jul 2024, 08:42
Originally posted by MathRevolution on 01 Sep 2019, 22:59.
Last edited by Bunuel on 19 Jul 2024, 08:42, edited 2 times in total.
Last edited by Bunuel on 19 Jul 2024, 08:42, edited 2 times in total.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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\(A=\{1, 2, 3, 4, 5\}\) is given. If there are \(1,2,…,n\) subsets of \(A\), let A1, A2 , A3 ,… , An be the sums of the elements of the corresponding subsets of \(A\). What is A1+ A2,+ A3+… + An?
\(A. 200\)
\(B. 210\)
\(C. 230\)
\(D. 240\)
\(E. 250\)
\(A. 200\)
\(B. 210\)
\(C. 230\)
\(D. 240\)
\(E. 250\)
08 Sep 2019, 16:16
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tgubbay1
The subsets of \(A = \{1,2,3,4,5\}\) are as follows.
\(S_1 = \{\}\),
\(S_2 = \{1\}\),
\(S_3 = \{2\}\),
\(S_4 = \{3\}\),
\(S_5 = \{4\}\),
\(S_6 = \{5\}\),
\(S_7 = \{1,2\}\),
\(S_8 = \{1,3\}\),
\(S_9 = \{1,4\}\),
\(S_{10} = \{1,5\}\),
\(S_{11} = \{2,3\}\),
\(S_{12} = \{2,4\}\),
\(S_{13} = \{2,5\}\),
\(S_{14} = \{3,4\}\),
\(S_{15} = \{3,5\}\),
\(S_{16} = \{4,5\}\),
\(S_{17} = \{1,2,3\}\),
\(S_{18} = \{1,2,4\}\),
\(S_{19} = \{1,2,5\}\),
\(S_{20} = \{1,3,4\}\),
\(S_{21} = \{1,3,5\}\),
\(S_{22} = \{1,4,5\}\),
\(S_{23} = \{2,3,4\}\),
\(S_{24} = \{2,3,5\}\),
\(S_{25} = \{2,4,5\}\),
\(S_{26} = \{3,4,5\}\),
\(S_{27} = \{1,2,3,4\}\),
\(S_{28} = \{1,2,3,5\}\),
\(S_{29} = \{1,2,4,5\}\),
\(S_{30} = \{1,3,4,5\}\),
\(S_{31} = \{2,3,4,5\}\),
\(S_{32} = \{1,2,3,4,5\}\).
The half of 32 sets, 16 sets have an element 1, 2, 3, 4 and 5 respectively.
When we calculate \(A_1 + A_2 + \cdots + A_{32}\), 1 happens 16 times, 2 happens 16 times, ... and 5 happens 16 times.
Thay's why we have \(A_1 + A_2 + \cdots + A_{32} = 16*1 + 16*2 + ... + 16*5 = 16(1+2+3+4+5) = 16*15 = 240\).
Answer: D
General Discussion
01 Sep 2019, 23:00
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=>
\(A\) has \(2^5 = 32\) subsets, so \(n = 32\). Each element of \(A\) is contained in \(2^{5-1} = 2^4 = 16\) subsets of \(A\) (We fix one element and find the number of subsets of the remaining elements). So, A1 + A2 + A3 + … + A32 = \(16*(1+2+3+4+5) = 240.\)
Therefore, the answer is D.
Answer: D
\(A\) has \(2^5 = 32\) subsets, so \(n = 32\). Each element of \(A\) is contained in \(2^{5-1} = 2^4 = 16\) subsets of \(A\) (We fix one element and find the number of subsets of the remaining elements). So, A1 + A2 + A3 + … + A32 = \(16*(1+2+3+4+5) = 240.\)
Therefore, the answer is D.
Answer: D
