keibee82 said:

hmmm we can have 222, 333 or 444. why is the possibility for second

spot 2 and third spot 1?

Given that the numbers involved are exactly 2, 3, and 4 (ie. ignoring the possibilities of 222, 333, and 444). The possible permutations computed like this (or at least think of it in this way...)...

There are 3 spots.... _, _, _ ....one for each number (2, 3,

**OR** 4). And there are 3 numbers (2, 3, and 4). In the first spot, we havn't placed in any number yet. Thus, we can either put in a 2, a 3, or a 4. In other words, we can place THREE possible numbers in the first spot.

Once we place a single digit in the first spot (doesn't matter if it's a 2, a 3, or a 4), we have TWO remaining numbers. These two remaining numbers MUST go in the remaining two spots. Thus, for the second spot, we can place in either of TWO possible numbers.

Once we've placed numbers for the first and second spot, there is only ONE remaining number, and ONE remaining spot. There is only ONE way to place this remaining number.

Thus, overall ...ASSUMING WE'RE USINGS THE NUMBERS 2, 3,

**AND ** 4, the possible permutations are ...3x2x1=6.

Now, add 3 (for 222, 333, and 444).

This is the precise way to get the answer. But my way is the quick and dirty way to getting the answer. Get the answer the quickest way possible, preferably with the least amount of complex thinking! I was on my high school math team! I was top 20 in the state! ...JUST GET THE JOB DONE!!!