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How many different fractions can be made up of the numbers 3, 5, 7, 11, 13, 17 so that each fraction contains 2 different numbers? How many of them will be the proper fractions?

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    $\begingroup$ Hi, welcome to MSE. Please provide some attempts or context to the problem, so that the community can better help you. For the second question, a hint is that the numbers are all prime. $\endgroup$ Commented Jul 2, 2020 at 15:55
  • $\begingroup$ This is not clear. How does a fraction "contain" two different numbers? Suppose the numbers were just $3,5$. What would the answer be then? What about $3,5,7$? $\endgroup$
    – lulu
    Commented Jul 2, 2020 at 15:55
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    $\begingroup$ I guess the OP means that one number is chosen to be the numerator and one number chosen to be the denominator. $\endgroup$ Commented Jul 2, 2020 at 15:57
  • $\begingroup$ Alwats add your attempt with your question thank you. $\endgroup$ Commented Jul 2, 2020 at 16:06
  • $\begingroup$ $2{6\choose 2},\,{6\choose 2}$ @ValeriiaHordiienko .Consider the table $$\begin{array}{c|cccccc} &3&5&7&11&13&17\\ 3&\frac33&\frac53&\frac73&\frac{11}3&\frac{13}3&\frac{17}3\\ 5&\frac35&\frac55&\frac75&\frac{11}5&\frac{13}5&\frac{17}5\\ 7&\frac37&\frac57&\frac77&\frac{11}7&\frac{13}7&\frac{17}7\\ 11&\frac3{11}&\frac5{11}&\frac7{11}&\frac{11}{11}&\frac{13}{11}&\frac{17}{11}\\ 13&\frac3{13}&\frac5{13}&\frac7{13}&\frac{11}{13}&\frac{13}{13}&\frac{17}{13}\\ 17&\frac3{17}&\frac5{17}&\frac7{17}&\frac{11}{17}&\frac{13}{17}&\frac{17}{17} \end{array}$$. For 1. you take all except the diagonal, $\endgroup$ Commented Jul 2, 2020 at 16:06

2 Answers 2

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The numbers of fraction which have two different numbers according to @Benjamin wang = 6C2×2!=30. Now you can find the solution of next question. To be proper fraction I can't take the value 2! So, the answer of next question=6C2=15

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Note that all the numbers are mutually coprime integers and therefore there cannot be cancellations in the fractions. Then you only need to count the number of possible pairs.

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