From 4 oranges, 3 bananas and 2 apples, how many selections of 5 pieces of fruit can be made taking at least one of each kind?

Let's start by listing the possibilities by type of fruit. You have to take one of each kind, so the only options are:
3 oranges, 1 banana, 1 apple
2 oranges, 2 bananas, 1 apple
2 oranges, 1 banana, 2 apples
1 orange, 3 bananas, 1 apple
1 orange, 2 bananas, 2 apples

Now for each option, calculate the number of ways you can select individual fruits.
For the first one: (4 ways to select 3 oranges) x (3 ways to select 1 banana) x (2 ways to select 1 apple) = 24 ways to select 3 oranges, 1 banana, 1 apple.

There are 4 + 3 + 2 = 9 pieces of fruit.
No. of ways to select 5 pieces of fruit without restriction = ₉C₅

In case of no orange, select 5 pieces of fruit from the other 5 pieces of fruit (3 bananas and 2 apples), i.e. ₅C₅.
In case of no banana, select 5 pieces of fruit from the other 6 pieces of fruit (4 oranges and 2 apples), i.e. ₆C₅.
In case of no oranges, select 5 pieces of fruit from the other 7 pieces of fruit (4 oranges and 3 bananas), i.e. ₇C₅====
The answer:
No. of selections that take at least one of each kind
= ₉C₅ - (₅C₅ + ₆C₅ + ₇C₅)
= 126 - (1 + 6 + 21)
= 126 - 28
= 98

1 orange, 2 apples and 2 bananas
1 orange, 1 apple and 3 bananas
2 oranges, 2 apples and 1 banana
2 oranges, 1 apple and 2 bananas
3 oranges, 1 apple and 1 banana
.. AND THAT'S IT !
You cannot have 4 oranges in a selection of 5 pieces of fruit.
JUST 6 SELECTIONS

There are two correct answers to this math word problem.
The first one is grouping by type of fruit and irrelevant which particular orange, banana, or apple. Since you must have one of each, one of each gets removed.
You must then have 2 picks from 3 oranges, 2 bananas, and an apple.
Since only 2 picks, it is also the same as 2 oranges, 2 bananas, and 1 apple.
That gets you to an answer of "5" 
o o o b a   3 oranges, 1 banana, 1 apple 
o b o b a   2 oranges, 2 bananas and 1 apple
o a o b a 2 oranges, 1 banana, and 2 apples
b b o b a 1 orange, 3 bananas, and 1 apple
b a o b a 1 orange, 2 bananas, and 2 apples

However, if each piece of fruit is unique, and that is a "selection" the choices are much higher and the combinations is correct.
There are 157 selections of unique fruit pieces with at least one of each type.
Oranges 1-2-3-4-5, Bananas 6-7-8, apples 9-0
01236, 01237, 01238, 01246, 01247, 01248, 01256, 01257, 01258, 01267, 01268, 01269, 01278, 01279, 01289, 01346, 01367, 01368, 01369, 01378, 01379, 01389, 01467, 01468, 01469, 01478, 01479, 01489, 01567, 01568, 01569, 01578, 01579, 01589, 01678, 01679, 01689, 01789, 02347, 02348, 02356, 02357, 02358, 02367, 02368, 02369, 02378, 02379, 02389, 02456, 02457, 02458, 02467, 02468, 02469, 02478, 02479, 02489, 02567, 02568, 02569, 02578, 02579, 02589, 02678, 02679, 02689, 02789, 03456, 03457, 03458, 03467, 03468, 03469, 03478, 03479, 03489, 03567, 03568, 03569, 03578, 03579, 03589, 03678, 03679, 03689, 03789, 04567, 04568, 04569, 04578, 04579, 04589, 04678, 04679, 04689, 04789, 05678, 05679, 05689, 05789, 12369, 12379, 12389, 12469, 12479, 12489, 12569, 12579, 12589, 12679, 12689, 12789, 13679, 13689, 13789, 14679, 14689, 14789, 15679, 15689, 15789, 16789, 23469, 23479, 23489, 23569, 23579, 23589, 23679, 23689, 23789, 24569, 24579, 24589, 24679, 24689, 24789, 25679, 25689, 25789, 26789, 34569, 34579, 34589, 34679, 34689, 34789, 35679, 35689, 35789, 36789, 45679, 45689, 45789, 46789, 56789