The fruits are individual items, so apples are distinguishable from each other and pears are distinguishable from each other.

Let the purple basket have $k$ total fruits. Since purple must have more fruits than orange, and each basket must have at least 1 apple and 1 pear:
Total fruits = 8
Each basket needs at least 2 fruits, so purple can have k=5 or k=6 .


Case 1: Purple has 5 fruits
Then orange has 3 fruits.
If purple has $a$ apples and $5-a$ pears, then:

Purple needs at least 1 apple and 1 pear: \(1 \leq a \leq 4\)
Orange needs at least 1 apple and 1 pear:
Orange apples \(=5-a \geq 1 \Rightarrow a \leq 4\)
Orange pears \(=3-(5-a)=a-2 \geq 1 \Rightarrow a \geq 3\)


So a=3 or 4 .
a=3 : choose 3 of 5 apples and 2 of 3 pears:


\(\binom{5}{3}\binom{3}{2}=10 \cdot 3=30\)


a=4 : choose 4 of 5 apples and 1 of 3 pears:


\(\binom{5}{4}\binom{3}{1}=5 \cdot 3=15\)




Case 1 total: 30+15=45

Case 2: Purple has 6 fruits

Then orange has 2 fruits. Orange must have 1 apple and 1 pear:
Choose 1 apple for orange: \( \binom{5}{1}=5 \)
Choose 1 pear for orange: \(\binom{3}{1}=3 \)


Case 2 total: \(5 \cdot 3=15 \)

Total ways:


45+15=60



60