For the guy who don't have an idea on the subject see my little tutorial.

What is Permutations & Combinations?
But Don't understand How to do ones like these?

What is the number of permutations that you can have by taking 3 letters at a time and the first letter to be m or n?

Very easy when you get it right. Just notice the 2nd tree of your first post.

We had 3 letters there {a, b,c}. We were asked to get the permutations and we had 6.

Note the first letter on them. We had 2 'a's, 2 'b's and 2 'c's as the first letter.

So to get the number of permutations which starts with letter 'a', you need to divide the result by n (n=3).

So if you were asked to get the number of permutations that starts with letter a or b. It would be,

(

^{3}P

_{3} / 3) * 2 = (6 / 3) * 2 = 4

Okay. Let's consider your question.

Our set S is {k, l, m, n, o}. n= 5. We were asked to pick 3 letters at a time means k = 3.

^{5}P

_{3} = 5! / (5 - 3)! = 5! / 2! = 5 x 4 x 3 = 60

On these 60 permutations, we have equal number of 'k', 'l', 'm', 'n' and 'o' letters.

Number of permutations from 1 letter is, 60 / 5 = 12.

So, Number of permutations from 2 letters is, (60 / 5) x 2 = 24

I'll make the question a bit difficult now

Let's say we were given the same set {k, l, m, n, o}. Can you tell me the number of permutations that you can get when you chose 3 letters with first letter 'm' and the second letter not 'm'. Just evaluate your answer and see my answer.

^{5}P

_{3} = 60

To have first letter m = 60 / 5 = 12

When you select m once, you can't select it again. So the answer is obviously 12.

Okay another easy one.

From the same set {k, l, m, n, o}, you were asked to pick 3 numbers where the first letter to be 'k' and the second to be 'n'.

^{5}P

_{3} = 60

First letter to be 'k' = 60 / 5 = 12

Out of these 12, the second letter also repeats equal times. Since we have already selected the first letter, we only have 4 letters now. So each letter repeats 12 / 4 times. So the answer is 3.

See how easy the permutations are. You just need to read the question carefully. That's all. Combinations are even more easier.