Calculation of oxidation states:

• $[Cr(C_{6}H_{6})_{2}]$: Benzene is a neutral ligand, so $Cr$ is in $0$ oxidation state.
• $[Cr(H_{2}O)_{6}]Cl_{3}$: $H_{2}O$ is neutral and there are 3 $Cl^{-}$ ions, so $x + 0 = +3 \Rightarrow x = +3$.
• $K_{2}[Cr(CN)_{2}(O)_{2}(O_{2})(NH_{3})]$: $K$ is $+1$, $CN$ is $-1$, $O$ (oxo) is $-2$, $O_{2}$ (peroxo) is $-2$, and $NH_{3}$ is $0$. $2(+1) + x + 2(-1) + 2(-2) + (-2) + 0 = 0 \Rightarrow x = +6$.

Maximum range occurs at $45°$: $R_{\max} = \dfrac{u^2}{g}$

$u^2 = R_{\max}\times g = 16000 \times 10 = 160000$

$u = \sqrt{160000} = \mathbf{400\text{ m/s}}$

The time period of a pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$, so $L \propto T^2$.

• Initial length: $L = kT^2$
• New length: $L + \Delta L = k T_M^2$
• Strain: $\frac{\Delta L}{L} = \frac{T_M^2 - T^2}{T^2} = (\frac{T_M}{T})^2 - 1$
• From $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{Mg/A}{\Delta L/L}$
• Therefore, $\frac{1}{Y} = \frac{\Delta L/L}{Mg/A} = [(\frac{T_M}{T})^2 - 1]\frac{A}{Mg}$

Multiples of 4 in $\{1,\ldots,20\}$: $\{4,8,12,16,20\}$ — 5 elements. Their images must be multiples of 3 in $\{1,\ldots,20\}$: $\{3,6,9,12,15,18\}$ — 6 elements.

Assign the 5 multiples-of-4 to 5 of the 6 multiples-of-3 (injectively, since $f$ is onto): $6 \times 5 \times 4 \times 3 \times 2 = 6!/(6-5)! = 720$... but we need surjection on all 20 elements.

The remaining 15 domain elements map to the remaining 15 codomain elements (after fixing the 5 images above): $(15)!$ ways. But 1 multiple-of-3 is unused by the multiples-of-4, so it must be covered by remaining 15 elements.

Total $= 6! \times (15)! = (15)! \times 6!$