Combinations & Permutations Comments
Comments

Actually glad to have cleared that confusion.

I got a little confused about N as possibilities and E as Dials. But I still got the correct answer.

There are 10 ways to set each of the 3 dials... so the first combination would be 000, and the last one, 999.

why is 10^{3} incorrect?

@Ian, in this question you need to enter the actual answer.

What's gamma(13)?

12 factorial

Nice!

I like where you are going with these challenges, but would like to see more examples. Some of the working can be tricky to understand. One way of doing this would be to show 1 or two examples with failure to show how not do calculate some of the challenges.
Having a good understanding of what not to do can be just as powerful or greater to the student then just finding the correct answer.

@Joshua Thank for your feedback. I think that's a good idea, and I will look into doing it. However, I'll want to balance it with making it easy for people to look up how to do a specific topic without being distracted by the incorrect approach.

Great explanation, but isn't 7! = 5040? Do you mean 6! = 720?

It's 10! / 7! = 720
Sorry the division line wasn't showing clearly. 
AZ 26 letters.

I was stupid to think that there were 27 alphabets .

Confused (again =p)! We were learning about permutations and the answer seems to be exponents?
Based on the previous nodes I would have assumed 9 * (26!/23!) * (10!/7!) would be the way to work this one out? Did I get lost somewhere along the way?

These are practice challenges for what was covered until now, not just the last page. "Permute" is only used when one less element becomes available each time, but here you can reuse the same characters. See Permutations with Repetition for more.

I don't understand this one:
it shouldn't be 9!, because the first time you can choose among 9, but the second input is a choise among 8, because you can't reselect the one you are on
should't it be
(9*8*8*8)+(9*8*8*8*8)+(9*8*8*8*8*8)+(9*8*8*8*8*8*8)+(9*8*8*8*8*8*8*8)+(9*8*8*8*8*8*8*8*8) ? 
@Jacopo, 9! would give you the number of 9dot permutations, but you also want the total number of 4,5,6,7, and 8 dot permutations. After picking 2 dots, you no longer have 8 dots left...

Tricky!  There are 8 different 'spots' to put/arrange a tiger along a line of 7 Bears> x B x B x B x B x B x B x B x
4 tigers for 8 possible spots. 
(10 choose 7) * (8 choose 4) * ((6 choose 3)  4)

I don't agree with the answer, ex: if we have 3 N on string, the combination of the balls is NC3+NC2+NC1+NC0 because NC3 is BBB, NC2 are BBW,BWB,WBB, NC1 are WWB,WBW,BWW, and NC0 is WWW... nb: W is (NB)... I think the correct formula is like the cross genetics of mendel in Biology, 2^{N...}

@Michael If you had 2 choices each time, there would be 2^{N} possible permutations. But you're given a set number of black and white beads, so eventually you will only be able to use 1 color.
You could make a tree to determine the possibilities for a specific example, but a simpler solution is to use Choose.
Chillex
Aug 9, 10:12 PMNice explanation. Clear, straightforward and sufficient materials.