Rom

People like you are such assholes.

Fine.  I'm gone.

\(27=3^3 = 3 \cdot 9\\ 36=2^2 3^2 = 4 \cdot 9\\ \text{So if teams have 3 girls and 4 boys you can have 9 teams}\)

\(\text{I don't think you mean in order. I think you mean how many distinct arrangements are there}\\ \text{1) there are 11 total distinct balls. There are 11! = 39916800 arrangements}\\ \text{2) there are 4 identical balls so we have 4! identical arrangements. This gives a total of $\dfrac{11!}{4!} = 1663200$ arrangements}\\ \text{3) $\dfrac{11!}{7!4!} = 330$ arrangements}\)

\(p=\dfrac{\dbinom{3}{2}\dbinom{8}{1}}{\dbinom{10}{3}} + \dfrac{1}{10}-\dfrac{\dbinom{3}{2}\dbinom{8}{1}}{\dbinom{10}{3}} \cdot \dfrac{10}{10}= \dfrac{7}{25}\)

\(P[\text{player 1 gets A, 2, 3}]= \dfrac{\dbinom{4}{1}\dbinom{4}{1}\dbinom{4}{1}}{\dbinom{12}{3}} = \dfrac{16}{55}\\ P[\text{player 2 gets A, 2, 3|player 1 did}] = \dfrac{\dbinom{3}{1}\dbinom{3}{1}\dbinom{3}{1}}{\dbinom{9}{3}}=\dfrac{9}{28}\\ P[\text{player 3 gets A,2,3|1 and 2 did}] = \dfrac{\dbinom{2}{1}\dbinom{2}{1}\dbinom{2}{1}}{\dbinom{6}{3}}= \dfrac{2}{5}\\ p= \dfrac{16}{55}\dfrac{9}{28}\dfrac{2}{5} = \dfrac{72}{1925}\)

\(z^4 = -4 = 4e^{i(2k+1)\pi},~k\in \mathbb{Z}\\ z = \sqrt{2}e^{i(2k+1)\pi/4},~k=0,1,2,3\\ z = 1+i,~-1+i,~-1-i,~1-i\)

what I don't think is right is that all the animals are distinct.

I suspect the problem want's them identical by type.

\(\text{The number of components that fail has a binomial distribution, $n=12,~p=0.2$}\\ P[k>3]=1-P[0]-P[1]-P[2]-P[3] = \\ 1 - \sum \limits_{k=0}^3 \dbinom{12}{k}(0.2)^k(0.8)^{12-k} = 0.205431\\ \)

\(P[E]=2P[O]\\ P[E]+P[O]=1\\ P[E]=\dfrac 2 3,~P[O]=\dfrac 1 3\\ P[1]=P[3]=P[5]=\dfrac 1 3 \cdot \dfrac 1 3 =\dfrac 1 9\\ P[2]=P[4]=P[6]=\dfrac 1 3 \cdot \dfrac 2 3 = \dfrac 2 9\\ P[roll < 4]=P[1]+P[2]+P[3] = \dfrac 1 9 + \dfrac 2 9 + \dfrac 1 9 = \dfrac 4 9\)

\(\text{clearly all the $a$'s must be such that $0 \leq a_k \leq n$}\\ \text{given any set of $k$ integers there is exactly one unlabelled ordering of them satisfying above}\\ \text{so we can choose $k$ integers $0\leq a_k \leq n$ and sort them, and label them accordingly}\\ \text{there are $N=(n+1)^k$ different sets of $k$ integers in the range $[0,n]$}\)

\(p = \dfrac{\dbinom{9}{3}\dbinom{52}{0}}{\dbinom{61}{3}} = \dfrac{42}{17995} \approx 0.002334 = 0.2334\%\\ \text{So yeah, pretty small probability}\)

A) \(\text{There are 3! orderings of the first 3 streets and then 5! ways to arrange the rest of them}\\ \text{There are a total of 8! orderings of the streets}\\ p = \dfrac{3!5!}{8!} = \dfrac{6}{336} = \dfrac{1}{56}\)

B) \(\text{Now there is only 1 ordering of the first 3 streets, and then 5! orderings of the rest}\\ p = \dfrac{5!}{8!} = \dfrac{1}{336}\)

You're always going to have at least 1 emu, chicken pair.  So we make that and reduce the problem.

There are 7*5 = 35 ways to make that pair.

Now he have 4 chickens, 4 donkeys, and 6 emus to pair off.

We must pair off emus with an equal number of chickens and donkeys, otherwise we'll have a pair of the same animal left at the end.

We must pair off all the emus so we must have 3 emu/chicken pairs, and 3 emu/donkey pairs.  Then we will have 1 chicken/donkey pair left.

There are 6C3 *3! * 4C3*3! ways to make the emu/chicken pairs.

Then there are 3C3*3! * 4C3*3! ways to make the emu/donkey pairs.

We then have 1 chicken and 1 donkey left that pair up.

That makes (including the 35 ways to make the first emu/chicken pair) 14515200 different arrangements.

Look at it another way.

I have 6*4 choices for the first emu/chicken pair, then 5*3, then 4*2.

I then have 3*4 choices for the first emu/donkey pair, then 2*3, then 1*2

This gives me

35 * 6*4*5*3*4*2*3*4*2*3*1*2 = 14515200

\(\text{t is Tom, f is father}\\ (t-5)=\dfrac{f}{3}\\ (t+5)=\dfrac f 2\\ \text{add these}\\ 2t = \dfrac 5 6 f\\ t = \dfrac{5}{12}f\\ \dfrac{5}{12}f - 5 = \dfrac f 3\\ \dfrac{f}{12}=5\\ f = 60,~t=25\)

\(\text{One simple way to do this is to change the constant on the right hand side until $(3,6)$ solves the equation}\\ 4(3)-3(6) = 12-18 = -6\\ 4x-3y=-6\\ \text{is parallel to $4x-3y=21$ and passes through $(3,6)$}\)

\(P[\text{5 games}] = P[\text{5 games | A wins}]+P[\text{5 games | B wins}]\)

\(P[\text{5 games | A wins}]=p^4 (1-p)\\ P[\text{5 games | B wins}]=p (1-p)^4\\ P\text{5 games}=p^4(1-p)+p(1-p)^4 = \\ p(1-p)[p^3+(1-p)^3]\)