Keith,

Suppose that your 13 rocks are different and you want to select 12 of them to put in a bucket - you don't care how they are piled up in the bucket, just which 12. A simple approach here is to think of which rock is left behind = there are 13 possibilities for the rock left behind therefore there are 13 ways to pick 12 for the bucket.

For y rocks and x to go into the bucket it is much more difficult - we need the number of ways to choose x from y distinct objects. We usually use yCx for this. If you need this amount of detail then write back.

yCx is ^{y(y-1)(y-2)...(y-x+1)}/_{x(x-1)(x-2)(x-3)...(3)(2)(1)}

Penny

Keith wrote back.

Thanks for the answer to my question yesterday however i do have a followup question.
 
 i have a bucket that can hold 12 rocks,if i have 13 rocks there are 13 possible combinations of rocks i can put in the bucket without having the exact same 12 rocks in the bucket.
 
if i use the same bucket that can only hold 12 rocks and i have 14 rocks how many possible combinations of rocks can i put in the bucket without having the exact same 12 rocks in the bucket.
 
what if i have 15 or 16,17,18etc...rocks how many possible combinations of rocks can i put in the bucket without having the exact same 12 rocks in the bucket.     

thanks very much for your help 

If you had 14 think about the 2 you're leaving out; in how many ways can you select them? You can choose the 1st in 14 ways and the 2nd in 13 ways but we need to be careful since the order is not important here - what does that mean? Well if you line up two distinct objects you can do it in 14 13 ways but if you are just picking two things and not lining them up we need to divide by the number of ways we can line up two objects which is 2 1. So the answer you want is 14 13 12/(3 2 1). And so on for other number of rocks.

Penny
                                                                        

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