Combinatorial Straight Lines

The number of ways of selecting 2 numbers A and B from 6 digits is $P_{2}^6 = 30$ ways. But in the above computation, the straight lines corresponding to the following cases are the same straight line which have been overcounted :

$(1)$ $A=0$ and $B = 1,2,3,6,7$

$(2)$ $B=0$ and $A = 1,2,3,6,7$

$(3)$ $A=1, B=2$ and $A=3, B=6$

$(4)$ $A=1, B=3$ and $A=2, B=6$

$(5)$ $A=2, B=1$ and $A=6, B=3$

$(6)$ $A=3, B=1$ and $A=6, B=2$

Hence, the number of distinct straight lines is $30 - 2 \times 4 - 4 = 18 ways$

Calculate first the cases that zero is not a or b, so we have 5C2 *2 =20 , we should multiply by 2 because a and b do not have the same role , so if there are a and b that a/b or b/a , and the a/b or b/a and 1 are both in the set {1,2,3,6,7} (1 is already in this set),we should subtract the (1,a/b) and (a/b,1) , or the (b/a,1) and (1,b/a) from the 20 calculated above (5C2 *2 =20) , here we have 3/6 and 2/6 , so 20 -2-2 =16 , and finally if a=0 all lines for b in the set {1,2,3,6,7} are the same,so 16+1=17 , and if b=0 the same thing ,so 17+1=18.