The question is:
From 4 oranges, 3 bananas and 2 apples, how many selections of 5 pieces of fruit can be made, taking at least 1 of each kind?
My attempt at this is as follow:
- Choose 1 orange from 4:$\binom{4}{1}$ = 4 ways.
- Choose 1 banana from 3:$\binom{3}{1}$ = 3 ways.
- Choose 1 apple from 2: $\binom{2}{1}$ = 2 ways.
- Choose 2 fruits from the remaining (3 oranges, 2 bananas, 1 apple): $\binom{6}{2}$ = 15
Multiplying these together: $4 \times 3 \times 2 \times 15 = 360$ which is incorrect because this exceed the total number of selections of any 5 fruits without any restriction: $\binom{9}{5}$.
However, other solutions that I have found either:
- Add up different combinations of the fruit: Is there a shortcut to this combination problem?
- Take the total number of selections, then subtract each selection that fails to have at least one of a kind.
Both of these solutions lead to 98.
I understand the 2 correct solutions but I am being torn up not being able to find out the flaw of my own attempt at the solution. Can someone please point out the difference in my incorrect solution and the correct ones? Why is it wrong?
Thank you so much in advance.
Best Answer
Your answer ($360$) is incorrect because of double counting.
Let's say the four oranges are labeled $O_1$, $O_2$, $O_3$, and $O_4$.
Now let's consider two possibilities following the steps you included.
1. We choose $O_2$ in the first step and $O_4$ in the last step.
2. We choose $O_4$ in the first step and $O_2$ in the last step.
Assuming everything else is the same in the two cases, we end up with exactly the same fruits. But it is being counted twice.
In certain cases, the same outcome might get counted more than twice. That's why, your answer is way higher than the correct one.