Let S = {1, 2, 3, 5, 7, 10, 11}. The number o | vedclass

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(A) Let $S = \{1, 2, 3, 5, 7, 10, 11\}$. We classify the elements based on their remainder when divided by $3$: $R_0 = \{3\}$ (count $n_0 = 1$) $R_1 =

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$42$

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(A) Let $S = \{1, 2, 3, 5, 7, 10, 11\}$. We classify the elements based on their remainder when divided by $3$: $R_0 = \{3\}$ (count $n_0 = 1$) $R_1 = \{1, 7, 10\}$ (count $n_1 = 3$) $R_2 = \{2, 5, 11\}$ (count $n_2 = 3$) Let $a_n, b_n, c_n$ be the number of subsets of size $n$ with sum $\equiv 0, 1, 2 \pmod 3$ respectively. The generating function is $P(x, y) = (1+x)(1+xy)(1+xy^2)^3(1+xy)^3 = (1+x)(1+xy^2)^3(1+xy)^4$. Using the root of unity filter,the number of subsets with sum $\equiv 0 \pmod 3$ is $\frac{1}{3} [P(1, 1) + P(1, \omega) + P(1, \omega^2)]$. $P(1, 1) = 2^7 = 128$. $P(1, \omega) = (1+1)(1+\omega^2)^3(1+\omega)^4 = 2(-\omega)^3(-\omega^2)^4 = 2(-1)(\omega^8) = -2\omega^2$. $P(1, \omega^2) = (1+1)(1+\omega)^3(1+\omega^2)^4 = 2(-\omega^2)^3(-\omega)^4 = 2(-1)(\omega^4) = -2\omega$. Sum $= \frac{1}{3} [128 - 2(\omega^2 + \omega)] = \frac{1}{3} [128 - 2(-1)] = \frac{130}{3}$. Since we need nonempty subsets,we subtract the empty set (sum $0$): $\frac{130}{3} - 1 = 42.33$. Re-evaluating: The subsets are: $n_0=1, n_1=3, n_2=3$. Total subsets with sum $\equiv 0 \pmod 3$ is $43$. Excluding the empty set,the answer is $43 - 1 = 42$.

Let $S = \{1, 2, 3, 5, 7, 10, 11\}$. The number of nonempty subsets of $S$ such that the sum of all elements is a multiple of $3$ is $........$

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