Consecutive and adjacent permutations help

[Eagerissac]

So I'm stuck on two questions below related to permutations and order. I think I might've gotten a) but I'm at an absolute lost as to what I'm supposed to do for b). I wrote my attempts below and was hoping someone could confirm my answers or explain how to do them? I would appreciate a step by step thought process because I've looked up similar questions with just the formulas provided and I don't understand them.

The answers I got as well were infinite and I don't understand if that's what they should be?

Consider all permutations of the integers 1,…,1000.

a. Calculate the number of permutations in which the number 1 appears before 2, 2 appears before 3, and 3 appears before 4? (In other words, 1,2,3,4 appear in order, but not necessariIy consecutiveIy.)

1000!/4! because there are 1000 integers to shuffle plus the others that are 1, 2, 3, and 4. I got an infinite answer when I calculated it.

b. Calculate the number of permutations in which the numbers 1,2,3,4 appear in order but no two are adjacent?

I'm stumped on how to answer this one. My attempt was to treat 1, 2, 3, and 4 as a single element which gives 997! numbers but I don't know what to do next or why. I just have a rough time picturing it.

 

[Dr.Peterson]

You are right for (a), but the answer is not infinite, just huge. There is a huge difference between the two, and you should know better. (Did you call it infinite because your calculator said so?) It's value is about 1.6766135836545573897654268080125*10^2566.

It's a nice trick: There are 1000! permutations of all 1000 numbers; they can be organized in groups of 4! where 1, 2, 3, and 4 are in the same positions but permuted, and one of those will have them in increasing order. Therefore the total number is 1000!/4!

Now for (b), here is one approach you might try: Consider any permutation of the numbers 5, 6, ..., 1000. To make a permutation of all 1000 numbers in which 1, 2, 3, 4 are separated, you could choose any 4 of the spaces between [or around] the 996 numbers, and insert 1, 2, 3, 4 (in order) into them. How many ways can you do this?

 

[pka]

Consider all permutations of the integers 1,…,1000.
a. Calculate the number of permutations in which the number 1 appears before 2, 2 appears before 3, and 3 appears before 4? (In other words, 1,2,3,4 appear in order, but not necessariIy consecutiveIy.)
1000!/4! because there are 1000 integers to shuffle plus the others that are 1, 2, 3, and 4. I got an infinite answer when I calculated it.

b. Calculate the number of permutations in which the numbers 1,2,3,4 appear in order but no two are adjacent?
I'm stumped on how to answer this one. My attempt was to treat 1, 2, 3, and 4 as a single element which gives 997! numbers but I don't know what to do next or why. I just have a rough time picturing it.

You are correct in part a) HERE is the exact answer.

For part b) we need to use the numbers \(\displaystyle 5,6,7,\cdots 998,999,1000\) to separate \(\displaystyle 1,2,3,4\) from one another.
Those 996 remaining numbers create 997 places to put \(\displaystyle 1,2,3,4\) in that order but separated. \(\displaystyle 997!=\) HERE