To solve this problem, we need to calculate the number of ways to choose 4 pieces of jewellery such that at least 1 necklace and at least 1 bracelet are included.

We can use two methods to solve this:

Method 1:

1. Choose 1 necklace, 2 rings, and 1 bracelet: \(\binom{2}{1} \times \binom{8}{2} \times \binom{4}{1} = 2 \times 28 \times 4 = 224\)
2. Choose 1 necklace, 1 ring, and 2 bracelets: \(\binom{2}{1} \times \binom{8}{1} \times \binom{4}{2} = 2 \times 8 \times 6 = 96\)
3. Choose 2 necklaces, 1 ring, and 1 bracelet: \(\binom{2}{2} \times \binom{8}{1} \times \binom{4}{1} = 1 \times 8 \times 4 = 32\)
4. Choose 2 necklaces and 2 bracelets: \(\binom{2}{2} \times \binom{4}{2} = 1 \times 6 = 6\)
5. Choose 1 necklace and 3 bracelets: \(\binom{2}{1} \times \binom{4}{3} = 2 \times 4 = 8\)

\(Total = 224 + 96 + 32 + 6 + 8 = 366 ways.\)

Method 2:

Calculate the total number of ways to choose 4 pieces from 14 items (2 necklaces + 8 rings + 4 bracelets) and subtract the cases where there are no necklaces or no bracelets:

Total ways: \(\binom{14}{4} = 1001\)

Subtract cases with no necklaces (only rings and bracelets): \(\binom{12}{4} = 495\)

Subtract cases with no bracelets (only necklaces and rings): \(\binom{10}{4} = 210\)

Subtract cases with only rings: \(\binom{8}{4} = 70\)

Using inclusion-exclusion principle, we find:

\(1001 - (495 + 210 - 70) = 366\)

Thus, the number of possible selections is 366.