Please post a screenshot of the exact question. I am not sure whether the fruits are to be taken as identical or distinct e.g. whether the 4 oranges are identical or distinct. If nothing is given, I would be tempted to take them as identical.

In that case, I would split 5 into 3 groups with at least one each in the following ways:

{3, 1, 1} - 2 ways (3 oranges or 3 bananas)
{2, 2, 1} - 3 ways (1 orange or 1 banana or 1 apple)

So 5 ways.

But if the fruit pieces are distinct, then we have 9 distinct pieces of which we have to select 5 in 9C5 ways.

Out of these 9C5 ways, you are not allowed those in which one type of fruit is not taken.
No of ways in which oranges are not taken = 5C5
No of ways in which bananas are not taken = 6C5
No of ways in which apples are not taken = 7C5

So total acceptable ways = 9C5 - 5C5 - 6C5 - 7C5 = 126 - 1 - 6 - 21 = 98
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total combinations = (a + b + 3o), (a + 2b + 2o), (a + 3b + o), (2a + b + 2o), (2a + 2b + 1o)

possible numbers of combinations = (2c1 x 3c1 x 4c3) + (2c1 x 3c2 x 4c2) + (2c1 x 3c3 x 4c1) + (2c2 x 3c1 x 4c2) + (2c2 x 3c2 x 4c1)
= (2x3x4) + (2x3x6) + (2x1x4) + (1x3x6) + (1x3x4)
=24 + 36 + 8 + 18 + 12
= 98

You buy at least one of each kind.

Number of oranges you can choose from = 4 -1 = 3
Number of bananas you can choose from = 3 - 1 = 2
Number of apples you can choose from = 2 - 1 = 1
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Total number of fruits you can choose from = 6

As you already have 3 fruits (one orange, one banana, and one apple),
you need to pick 2 more fruits.

6C2 = 15

we will have to make cases and solve
4c3*3c1*2c1+4c2*3c2*2c1+4c1*3c3*2c1+4c1*3c2*2c2+4c2*3c1*2c2 ; 24+36+8+12+18 ; 98

VeritasKarishma Bunuel
Please help with this one.

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