A math conference is presenting a lecture series with six different lecturers. If Dr. Smith’s lecture depends on Dr. Jones’ lecture, so that Dr. Smith must be scheduled at some time after Dr. Jones, in how many orders can the six lectures be scheduled?

The are multiple ways to compute the order

To solve this

We pick two of the six slots for Jones and Smith (to be scheduled in that order);

So there are

6C2 = 6! / (4! 2!) = 15 ways to do that.

The other four lecturers can fill the other four slots in any order,

and there are 4! ways to do so,

and the total number of possibilities are

6C2(picking two slots) * 4! (Picking remaining 4 in any order) =

Total orders = 15 * 24

Total Orders = 360.

the line of

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )