1. We need the reactions that are undergone by both acetaldehyde, $\ce{CH3CHO}$, and acetone, $\ce{CH3COCH3}$.

2. Check each reaction one by one:

   A. Iodoform reaction

   • Iodoform reaction is given by compounds containing the group $\ce{CH3CO-}$ or those oxidizable to it.
   • Acetaldehyde $\ce{CH3CHO}$ gives iodoform reaction.
   • Acetone $\ce{CH3COCH3}$ is a methyl ketone, so it also gives iodoform reaction.
   • Hence, A is true for both.

   B. Cannizzaro reaction

   • Cannizzaro reaction is shown by aldehydes without $\alpha$-hydrogen.
   • Acetaldehyde has $\alpha$-hydrogens, so it does not undergo Cannizzaro reaction.
   • Acetone is a ketone and does not undergo Cannizzaro reaction.
   • Hence, B is false.

   C. Aldol condensation

   • Aldol condensation is shown by aldehydes/ketones having at least one $\alpha$-hydrogen.
   • Acetaldehyde has $\alpha$-hydrogen.
   • Acetone also has $\alpha$-hydrogen.
   • Hence, C is true for both.

   D. Tollen's test

   • Tollen's reagent oxidizes aldehydes, so acetaldehyde gives the test.
   • Ordinary ketones like acetone do not give Tollen's test.
   • Hence, D is false for both together.

   E. Clemmensen reduction

   • Clemmensen reduction reduces carbonyl compounds (aldehydes and ketones) to hydrocarbons using $\ce{Zn(Hg)/HCl}$.
   • Acetaldehyde and acetone both undergo this reaction.
   • Hence, E is true for both.

3. Therefore, the reactions undergone by both are: $A,\ C,\ E$

4. Match with options:

   • Option A: $C$ and $E$ only $\to$ missing $A$
   • Option B: $A, B, D$ only $\to$ wrong
   • Option C: $A, C, E$ only $\to$ correct
   • Option D: $B, C, D$ only $\to$ wrong

So the correct option is C.