[Combinatorics] Selections, Distributions, and Arrangements with Multiple Restrictions
Responses should utilize appropriate combinatorial formulas; according to the assignment writeup, the solutions should contain explainations not just answers. Thank you in advance.
Problem 1: Find the total number of ways the game show host can distribute 5000 points to the four distinct teams (A,B,C,D) such that the following hold true:
 Every team receives at least 250 points
 Team A receives at least 1000 points
 Team B receives at most 600 points
Problem 2: Suppose we wish to create an anagram of DRACONIS LUCIUS MALFOY. Counting both the 2 interior spaces and letters as characters, find the total number of ways to arrange the characters such that the following hold true:
 at least 2 of the vowels are consecutive
 the first A comes before the first U
 the last letter is O
Problem 3: Suppose you want to create a new password. The password may consist of uppercase letters, lowercase letters, numbers 09, and special characters. Assume there are 9 types of special characters. If the password must be 18 characters in length and repetition is allowed, find the probability of choosing a password such that each of the following conditions are met:
 it does not end with an uppercase letter
 it contains exactly one distinct special character that is not '@' and is not in the 3rd or 4th position
 it contains exactly 3 (single digit) odd numbers that are distinct
 the remainder of the open positions in the password consists of lowercase letters where repetition is allowed

For the 3), by "probability" you mean if we choose each character at random? And the second condition, does that mean that the 3rd and 4th positions can contain any character, even a special one?

By probability, the question is asking for the solution to be {the number of passwords meeting all criteria} / {the total number of passwords with no restrictions}. In this case, I believe the total number of nonrestricted passwords would be (10+26+26+9)^18 = 71^18 since there are 10 numbers, 26 lowercase letters, 26 uppercase, and 9 special characters to choose from.

I believe the 3rd and 4th positions cannot have a special character, and the entire restricted password can only have one distinct special character that is not '@' (so you're left with 91=8 special characters to choose from).

Some notes on the problems: For problem 2, the arrangement can start with an empty space. For problem 3, the last bullet point refers to all the other characters not mentioned in the previous bullet points. For each problem, the bullet points must be satisfied simultaneously.

Ah sorry, I think there are a couple small mistakes in problem 2: I'm also counting the anagrams that begin with a vowel but don't have two consecutive vowels. There are \binom{12}{7} of them (no empty box except the first and the last, so no empty boxes out of 8 boxes), just subtract this from the expression in the parentheses (before multiplying by 13! 7! / 8). The second mistake is that of course 2!2!2! = 8 (not 4), which I did twice; just divide everything by 4. Is it clear what's wrong?
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Ah sorry, I think there are a couple small mistakes in problem 2: I'm also counting the anagrams that begin with a vowel but don't have two consecutive vowels. There are \binom{12}{7} of them (no empty box except the first and the last, so no empty boxes out of 8 boxes), just subtract this from the expression in the parentheses (before multiplying by 13! 7! / 8). The second mistake is that of course 2!2!2! = 8 (not 4), which I did twice; just divide everything by 4. Is it clear what's wrong?

Thank you for the fast solution. I'll look over it soon, and get back to you. Hope you're having a relaxing day!

Ok! Just letting you know that I'm in the EU time zone, so I'll be available for roughly 4 more hours!

Are you not allowed to edit you answer? It's fine if you can't, I'm just curious. So... in your fix for problem 2, should the parenthetical be ( \binom{218}{8}  \binom{12}{8}  \binom{12}{7} ) ? Also, I am not sure what you mean by divide everything by 4. Sorry for my confusion.

No, I can't edit, unfortunately! Yes, the parenthetical is that one, and it's 13!/16 * 7!/2 * that thing, not 13!/8 * 7!, I stupidly miscomputed 2!2!2! as 4 instead of 8.

Okay, that helps. Just to make sure, your solution to problem 2 does cover the "at least two of the vowels are consecutive" restriction?

Yes, that's the parenthetical. The first term (\binom{218}{8}) are all the possible ways to put the consonants among the vowels, the second one (\binom{12}{8}) are the ones that have at least one consonant between any two vowels and start with a consonant, and the last one which I forgot (\binom{12}{7} ) are the ones that have at least one consonant between any two vowels and start with a vowel. If you take the difference you get the ones with at least two consecutive vowels.

Okay, perfect! I hope you get some rest. Thank you again. I'll let you know if anything else is unclear, but it all looks good at the moment.

Ok, just let me know if you have any question or you think anything is unclear!

I just manually calculated \binom{218}{8}  \binom{12}{8}  \binom{12}{7}, and I got 0... I don't think that is correct

Here's the actual values: 1287495792=0

If you get a moment, please go through Problem 2 again. I do think that most of it is right, but the current calculations are not giving accurate results.

It's not \binom{218}{8}  \binom{12}{8}  \binom{12}{7}, it's \binom{21}{8}  \binom{12}{8}  \binom{12}{7}, which is a lot bigger!

There's no 8 in the first term, and of course \binom{13}{8} = \binom{12}{8} + \binom{12}{7}, that's a special case of the Pascal identity.

Okay, I see now. Sorry for the confusion. These comment interactions don't really make this process easy! Thank you again.

I just realised that I told you that it was \binom{218}{8} so many times in the comments! Ah, I'm sorry, it was right in the original answer. The full answer is (13! * 7! / 32) * ((21 choose 8)  (13 choose 8)), notice that (13 choose 8) = (12 choose 8) + (12 choose 7). According to Google it's roughly 1.9831176e+17.

Yeah, it's not the best unfortunately!

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