How many different fractions can be made up of the numbers

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?

• 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. Commented Jul 2, 2020 at 15:55
• 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 $3,5,7$?
– lulu
Commented Jul 2, 2020 at 15:55
• I guess the OP means that one number is chosen to be the numerator and one number chosen to be the denominator. Commented Jul 2, 2020 at 15:57
• $2{6\choose 2},\,{6\choose 2}$ @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, Commented Jul 2, 2020 at 16:06

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

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.