We need to count how many of the given cations give a characteristic precipitate with potassium ferrocyanide, $\mathrm{K_4[Fe(CN)_6]}$.

In qualitative salt analysis, $\mathrm{K_4[Fe(CN)_6]}$ gives characteristic precipitates with certain metal ions:

• $\mathrm{Cu^{2+}}$ gives a chocolate brown precipitate of copper ferrocyanide.
• $\mathrm{Fe^{3+}}$ gives a deep blue precipitate (Prussian blue / ferric ferrocyanide type complex).
• $\mathrm{Zn^{2+}}$ gives a white precipitate of zinc ferrocyanide.
• $\mathrm{Ba^{2+}}$ gives a white precipitate of barium ferrocyanide.

Now check each ion one by one:

1. $\mathrm{Cu^{2+}}$ $\rightarrow$ gives characteristic precipitate. ✅
2. $\mathrm{Fe^{3+}}$ $\rightarrow$ gives characteristic precipitate. ✅
3. $\mathrm{Ba^{2+}}$ $\rightarrow$ gives characteristic precipitate. ✅
4. $\mathrm{Ca^{2+}}$ $\rightarrow$ no characteristic precipitate used in this test. ❌
5. $\mathrm{NH_4^+}$ $\rightarrow$ no precipitate. ❌
6. $\mathrm{Mg^{2+}}$ $\rightarrow$ no characteristic precipitate in this test. ❌
7. $\mathrm{Zn^{2+}}$ $\rightarrow$ gives characteristic precipitate. ✅

Total number of cations giving characteristic precipitate: $4$

Comparison with stored correct answer: