The similarity transformation gives $A=PD{P}^{-1}$. In the matrix power, each adjacent pair $P^{-1}P$ cancels to the identity matrix:This also follows by mathematical induction: the formula holds for $n=1$, and multiplication by $A=PD{P}^{-1}$ turns its $n$th instance into the $(n+1)$st.

Direct matrix multiplication givesThus $D$ is diagonal. The columns of $P$ are eigenvectors of $A$ with eigenvalues one and four.

The determinant of $P$ is $2\cdot 1-1(-1)=3$. The two-by-two inverse matrix formula yields

On the same sphere, the normal component is $(-y,x,0)\cdot (x,y,z)=-xy+xy=0$. This vector field is tangent to the sphere everywhere. Therefore its outward flux is

On the unit sphere, the outward unit normal is $\mathbf{n}=(x,y,z{)}^T$, so the normal component of the radial vector field is $\mathbf{F}\cdot \mathbf{n}=x^2+y^2+z^2=1$. The direct surface integral isHere $dS=\sin \theta d\theta d\varphi$ is the unit-sphere area element of a surface in spherical coordinates.

On the surface $y=1+z$, the integrand is $x+z+z^2$. Using the area element of a surface from part (i),

Parameterize the surface by $\mathbf{r}(x,z)=(x,1+z,z{)}^T$. Its two tangent vectors are $(1,0,0{)}^T$ and $(0,1,1{)}^T$, whose cross product is $(0,-1,1{)}^T$ of Euclidean norm $\sqrt{2}$. Thus the scalar area element of a surface isEquivalently, projection onto the $xz$-plane reduces the surface area by $|\mathbf{n}\cdot {\mathbf{e}}_y|=1/\sqrt{2}$.

Here $\alpha =4$ and $\beta =4$, so the reduced ordinary differential equation is $(D-2{)}^{2}u=0$. Its repeated characteristic root gives $u=(A+Bt)e^{2t}$. The first initial condition sets $A=0$. The original equation $u^{\prime }=3u-v$ gives $u^{\prime }(0)=-1$, hence $B=-1$.

Recover $v=3u-u^{\prime }$ to obtainDirect differentiation verifies $v^{\prime }=u+v$ and both initial conditions.

Differentiate the first ordinary differential equation and substitute the second:The original first equation also gives $bv=u^{\prime }-au$, soTherefore the requested coefficients areThey are the trace and determinant of the coefficient matrix. This derivation never divides by $b$ and is valid even if $b=0$. In that degenerate case the scalar second-order equation is a necessary consequence, while the original constraint $u^{\prime }=au$ must also be retained when reconstructing the coupled system.

For first-bin size $k$, choose its contents in $\left(\genfrac{}{}{0ex}{}{6}{k}\right)$ ways. Each choice arises from $k!(6-k)!$ permutations, obtained by ordering the two bins internally. The seven ordered size choices giveHere $0!=1$. Each row expresses the same binomial coefficient identity $\left(\genfrac{}{}{0ex}{}{6}{k}\right)k!(6-k)!=6!$. Even for $\{3,3\}$ the bins retain their first/second labels, so the number is twenty rather than ten; the braces denote bin sizes in the question's convention, not interchangeable bins.

The product isEvery permutation determines exactly one bin partition, and each such partition has exactly forty-eight internal orderings. Thus the fifteen classes partition the original sample space of $720$ arrangements, without overlap or omission.

Choose the unordered pair of letters in the second bin; the first bin is then determined. The binomial coefficient givesThe first and second bins are distinguished, so their roles are not interchangeable.

The first four letters may be permuted in $4!$ ways and the last two in $2!$ ways without changing this partition. The multiplication principle therefore gives

The second bin is forced to contain $A,B$, leaving the other four letters in the first bin. Since order within a bin is irrelevant, this gives exactlyset partition of the specified bin sizes.

Requiring exactly one condition means taking the symmetric difference of the two sets. Subtract their shared cases twice from the two individual counts:Equivalently, remove the twenty-four arrangements satisfying both conditions from the inclusive count $216$.

Let $E_5$ denote the set of arrangements with $E$ fifth, and $F_6$ those with $F$ sixth. Each contains $5!$ arrangements, while their intersection contains $4!$. The inclusion-exclusion principle gives

Fixing both specified positions leaves four letters to permute. Hence

The last letter has two permitted choices, and each leaves $5!$ permutations. The two classes are disjoint, so the count is

Fixing the last letter leaves five distinct letters to permute freely. Therefore the number is

By the multiplication principle, the successive places have six, five, four, three, two and one available letters. Thus the number of permutations is