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Consider an $8$-element set $X$ with elements $\{x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8\}$. How many subsets of $X$ contain $x_2$ and $x_3$ but not $x_5$? Please give an arithmetic expression if possible.

Any ideas guys? I have absolutely no clue how to calculate it. Your help would be much appreciated.

yoknapatawpha
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kiasy
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3 Answers3

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Hint: We need to make a subset of $\{x_1,x_4,x_6,x_7,x_8\}$ to keep $x_2$ and $x_3$ company.

André Nicolas
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Consider an 8-element set X with elements {x1, x2, x3, x4, x5, x6, x7, x8}. How many subsets of X contain x2 and x3 but not x5? Please give an arithmetic expression if possible.

x2 and x3 must be in the set and x5 must not be in our set. This means that x1, x4, x6, x7, and x8 can be either in or out of our set, giving us two possibilities per each of these elements. There are 5 elements with 2 options for each, so by the product rule, the solution is $2^5$.

okarin
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  • what would be the arithmetic function though ? – kiasy Dec 19 '13 at 02:42
  • $2^5$ = $2 + 2 + ... + 2 + 2$ where you add 2 to itself 16 times. – okarin Dec 19 '13 at 02:43
  • So is 2^5 the arithmetic function ? @user112825 – kiasy Dec 19 '13 at 02:45
  • I believe the arithmetic function you are looking for is one of the following: $2^5$, $2 * 2 * 2 * 2 * 2$, or $2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2$. Whatever it is, it will be equal to 32. – okarin Dec 19 '13 at 02:46
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For any element except for $x_2,x_3$ and $x_5$ we have to chose if the element is in or not (so that is 5 yes or no choices) By the rule of product there are $2^5=32$ ways to chose.

Asinomás
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