Expert: Steve Holleran - 3/9/2007

Question
How many 6-number combinations are possible with ping pong balls numbered 1~56 for the first 5 numbers and ping pong balls numbered 1~46 for the 6th number (for example: 1,15,25,35,55, and 45)? also, if I want to  eliminate a certain type of five-number combinations (e.g. consecutive numbers like 1,2,3,4,5 or 2,3,4,5,6... etc.),  how would I write a formula for it? Thanks in advance.


Answer
Hi Jane,

Well, I can help with the first part for sure.

Since it doesn't matter which order the balls come up, for the first 5 numbers, you want to calculate a combination of 56 items taken 5 at a time :   56 C 5 =  3,819,816.

Then, since you have 46 possibilities for the "mega" ball,
multiply this result by 46:

3,819,816 * 46 =175,711,536.  You can see why buying one ticket and expecting to win is a bit of a stretch!

I'm not sure of any formulas that would give you the result you're looking for in the second question, though.
My grasp of probability doesn't go that deep.  For the example you pose, though, there would be 42 groups of consecutive numbers, starting with 1,2,3,4,5 and ending with 42,43,44,45,46.

Hope this helped a little bit.

Steve Holleran