This puzzle also has two solutions. Try to find both of them.

The five animal breeders each breed one of birds, cats, dogs, fish, and monkeys.

Only one animal breeder breeds the animal they share a name with.

The person who breeds dogs shares a name with the animal that FitzMonkey breeds.

LeBird breeds birds.
Which of the following additional constraints would give this puzzle a unique solution?
answer was McDog doesn't breed cats or monkeys.
LeBird birds, VanCat dogs, McDog ish, DuFish monkeys, FitzMonkey cats

Find the number of ways of selecting n letters from 3n letters which contains 'n' a s , 'n' b s and the rest n letters are distinct from each other.Solution:"There are $k+1$ ways to select $k$ a's and b's, and then there are $\binom nk$ ways to choose the distinct letters you don't select. Thus the desired count is

\begin{eqnarray*}
\sum_{k=0}^n(k+1)\binom nk
&=&
\left[\frac\partial{\partial q}\sum_{k=0}^n\binom nkq^{k+1}\right]_{q=1}
\\
&=&
\left[\frac\partial{\partial q}\left(q(1+q)^n\right)\right]_{q=1}

A coin is tossed $m+n$ times $(m>n)$.Show that the probability of atleast $m$ consecutive heads is $\frac{n+2}{2^{m+1}}$. Soution:"The probability the first $m$ tosses are heads is $\dfrac1{2^m}$.

The probability that the $j$th is tails and the next $m$ are heads is $\dfrac1{2^{m+1}}$, providing that $1 \le j \le n$.

If $n \le m$ then these are disjoint events, so the probability any of them occurs is $\dfrac1{2^m}+ n \times \dfrac1{2^{m+1}} = \dfrac{n+2}{2^{m+1}}.$" Doubt:why we are letting $j$th to be tail, if we don't let then we may get head at that place but that's what we want "g…