In permutation problems, order MATTERS. These are problems like "There are 10 people in a group. One person will be selected as president, one person will be selected as vice president, and another for treasurer. How many different choices can be made to fill these positions?"

In a combination problem, order DOES NOT MATTER. These are problems like "There are 10 people in a group. 3 will be selected for a survey. How many different selections are possible?"

i see...

thank you

one more question:
there is a permutation formula: n!/(n-r)!
and a combination formula: n!/k!(n-k)!

im confusing those 2 ...i guess i have to think to myself if the order matters or not, and then just use the right formula?

I don't use a formula for permutations. Refer to the permutation problem where we have 10 people, and we want to know how many possibilities for the pres/vice-pres/treasurer there are. Here is how I would solve this:

There are 10 choices to fill the president position. Now there are 9 remaining choices to fill the vice-president position since 1 person is occupying the president position.. Finally, there are 8 remaining choices since 2 people are occupying the previous 2 positions. There are 10*9*8 = 720 different ways to fill these 3 positions.

The formula will give you 10!/(10-3)! = 10*9*8, but it seems way easier to write it down by thinking about it, then plugging into some formula.

Now let's look at the the situation where we have 10 people and are choosing 3 of them to participate in a survey. There is no distinction between the choices. if we choose people ABC or BAC, those 3 people are taking the same survey.

I use the formula for combination problems. It's just quicker. Here is the idea behind it. We choose 3 people. We have 10 choices for the first pick, 9 choices for the second pick, and 8 choices for the third pick just like the last scenario. But this is an overestimate. This gives us 10*9*8 = 720 choices, but like we previously said, if we choose people A, B and C in the order ABC or CBA it doesn't matter.

So how many different ways can we choose people A B and C in the same group? We have 3 choices for the first pick, 2 choices for the second pick, and 1 choice for the third pick for a total of 3*2*1=3!.

So when order doesn't matter, you can divide the permutation answer by the repetition: 720/6 = 120.

We define n choose k to be n!/[(n-k)!k!]. People denote this as nCk. But look at this. n!/(n-k)! in this situation gives us 10*9*8, and then dividing by a factor of k! gives us 10*9*8/3!.

My approach is when order matters, to just figure out how many choices there are for each slot, and multiply the number of choices for each slot. This way is very intuitive.

If order doesn't matter, I use nCk and crunch out the number really quickly. The intuitive way takes too long, and can get confusing.

tks bro.