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