[Combinatorics] Selections, Distributions, and Arrangements with Multiple Restrictions

Problem 1:

First we assign $1000$ points to team A and $250$ points to teams B, C, and D. We are left with $3250$ points to assign. Now we assign $0 \leq k \leq 350$ more points to team B. For each such $k$, we have $\binom{3250-k+3-1}{3-1} = \binom{3252-k}{2}$ ways to assign the remaining $3250-k$ points to the teams A, C, D.

(remember that there are $\binom{n+r-1}{r-1}$ ways to distribute $n$ objects in $r$ boxes)

The total number of ways is then \[ \sum_{k=0}^{350} \binom{3252-k}{2} = \sum_{k=0}^{3252} \binom{k}{2} -  \sum_{k=0}^{2901} \binom{k}{2}= \binom{3253}{3} - \binom{2902}{3} = 1,662,883,326 \]

Problem 2:


First we sort the letters (and replace the white space with X for convenience). We have AAIIOOUUYCCDFLLMNRSSXX (22 letters, 9 vowels, 13 consonants). We deal with consonants and vowels separately. The consonants have 2+2+2+2 repetitions, so we have $\frac{13!}{2! 2! 2! 2!}$ anagrams for just those. For the vowels, we know that the last one must be an O, so we have $\frac{8!}{2!2!2!} = 2 \cdot 7!$ ways to arrange the remaining letters, and in exactly half of these, the first A comes before the first U (if it doesn't, you can swap them), so we are left with $7!$ possibilities. Now, think about the spaces between the vowels as boxes (including the space before the first one, but not counting the space after the last O, since it must be the last letter - so 9 spaces), and the consonants as things to put inside. There are $\binom{n-1}{k-1}$ ways to put $n$ things in $k$ boxes leaving no box empty, and $\binom{n+k-1}{k-1}$ ways to put $n$ things in $k$ boxes if we allow for empty boxes. In our case, we need at least one empty box, so it's enough to take the difference between the two numbers, and so we have \[ \binom{13+9-1}{9-1} - \binom{13-1}{9-1} \] possible fillings. Multiplying everything together, we have \[ \frac{13!}{8} \cdot 7! \cdot \left( \binom{21}{8} - \binom{12}{8}\right) \] anagrams.

Problem 3:

Let's count how many valid passwords we have. We have to choose a special character among 8 (not 9 since @ is excluded), so 8 ways; three distinct odd numbers, so $5 \cdot 4 \cdot 3 = 60$ possibilities if we take the order into account, and fill the remaining 14 spaces with lowercase letters, repetitions allowed, so $26^{14}$ choices. There is no uppercase letter, so the first condition is automatically satisfied. Now we assign the position of the special character: it can't be the 3rd or the 4th, so that leaves $16$ possibilities. Next, we assign the positions of the odd digits: we have $\binom{17}{3}$ ways of choosing them among the 17 we have left available. The positions of the lowercase letters are now assigned. 

We can conclude that we have \[ 8 \cdot 60 \cdot 26^{14} \cdot 16 \cdot \binom{17}{3} \] valid passwords. The total number of passwords is just the number of available characters ($26+26+10+9=71$) raised to the power given by the length, so $71^{18}$ choices. The probability we want is the ratio of these two numbers (which is roughly $10^{-8}$ if my computation is correct).

There shouldn't be mistakes in the computations but that can always happen, I hope the arguments are clear though, if not just tell me and I'll try to explain better.