Rom

$\text{4 choices, 15 options for each choice}\\ N = 15^4 = 50625‬$

$\text{It sounds like your asking for the number of ways you can partition the integer 4}\\ \text{using the digits 1, 2, 3, and 4}\\ (1,1,1,1)\\ (2,1,1) \text{ and it's other 2 permutations}\\ (2,2)\\ (1,3) \text{ and it's other permutation}\\ (4)\\ \text{for a total of $1+ 3+1+2+1 = 8$}$

$\text{a) $p = \dfrac{\dbinom{12}{3}\dbinom{16}{2}\dbinom{18}{2}}{\dbinom{46}{7}}$}$

$\text{b) $P[\text{at least 2 red balls withdrawn}] = 1 - P[\text{0 or 1 red balls withdrawn}]$}=\\ 1 - \dfrac{\dbinom{12}{0}\dbinom{34}{7}}{\dbinom{46}{7}} - \dfrac{\dbinom{12}{1}\dbinom{34}{6}}{\dbinom{46}{7}}$

$\text{c) $P[\text{all balls the same color}]$}=\\ \dfrac{\dbinom{12}{7}+\dbinom{16}{7}+\dbinom{18}{7}}{\dbinom{46}{7}}$

$\text{d) $P[\text{3 red or 3 blue withdrawn}]$}=\\ \dfrac{\dbinom{12}{3}\dbinom{34}{4} + \dbinom{16}{3}\dbinom{30}{4}}{\dbinom{46}{7}}$

$\text{The least common multiple of 6 and 7 is 42}\\ \text{It will be 42 days}$

$\text{Any set of 4 distinct integers can be put in strictly ascending order}\\ \text{We can choose $\dbinom{13}{4}=715$ sets of distinct integers in $[0,12]$}$

$\text{b) is the same idea as (a) but now the integers don't have to be distinct}\\ \text{we have a few flavors of how the selected integers might appear}\\ \text{i) 4 distinct integers, ii) 1 pair, ii) 2 pairs, iv) 3 of a kind, v) 4 of a kind}\\ \text{For all these flavors there is only a single way they can be distinctly sorted in ascending order}\\ \text{so $N=\dbinom{13}{4}+13 \dbinom{12}{2} + \dbinom{13}{2}+13\cdot 12 + 13 = 1820$}\\ \text{where each of those terms corresponds to the number of each flavor that occurs}$

you can do (c)

$\text{a) is equivalent to putting 50 identical balls into 4 distinct boxes}\\ \text{This is the well known stars and bars problem}\\ N=\dbinom{50+4-1}{4-1} = \dbinom{53}{3} = 23426$

$\text{b) is essentially the same as (a) but now there is a 5th box}\\ N=\dbinom{50+5-1}{5-1} = 316251$

$\text{Using the hockey stick identity}\\ \sum \limits_{n=13}^{54}\dbinom{n}{13} = \dbinom{54+1}{13+1} = \dbinom{55}{14}$

I don't agree with the answers given.

For (1) we can treat each candy individually and multiply the results.

Using stars and bars we can distribute 8 candies among 4 children as

$N_L = \dbinom{8+4-1}{4-1} = \dbinom{11}{3} = 165\\ N_W=\dbinom{7+4-1}{4-1}=\dbinom{10}{3} = 120\\ N = N_L \cdot N_W = 19800$

For (2) we have to find the sum of the cases for the amount the twins get

$N_0 = \dbinom{6+3-1}{3-1} = \dbinom{8}{2} = 28\\ N_1 = \dbinom{4+3-1}{3-1} = \dbinom{6}{2} = 15\\ N_2 = \dbinom{2+3-1}{3-1} = \dbinom{4}{2} = 6\\ N_3 = 1\\ 1 + 6 + 15 + 28 = 50$

$0.\overline{1}_b = \\ \sum \limits_{k=1}^\infty \dfrac{1}{b^k} \\ \text{let $u = \dfrac 1 b$}\\ \sum \limits_{k=1}^\infty \dfrac{1}{b^k} = \sum \limits_{k=1}^\infty u^k = \\ \dfrac{u}{1-u} = \\ \dfrac{\frac 1 b}{1-\frac 1 b} = \\ \dfrac{1}{b-1}$

$\text{let the distance to the truck be L}\\ h = L \tan(10^\circ)\\ h = (L-800)\tan(24^\circ)\\ (L-800)\tan(24^\circ)=L \tan(10^\circ)\\ L = \dfrac{800 \tan(24^\circ)}{\tan(24^\circ)-\tan(10^\circ)}\\ h = \dfrac{800 \tan(24^\circ)}{\tan(24^\circ)-\tan(10^\circ)}\tan(10^\circ)$