# Why don't these two different methods of counting give the same result?

We have 4 bananas, 5 apples, 6 oranges. How many ways can we choose 7 fruits with at least 4 oranges? The straight forward method is to divide this into cases with 4, 5, or 6 oranges and then picking from bananas and apples, so that we have chosen 7 fruits. If we compute each case and add them up, we get: $$\binom 64\binom 93+\binom 65\binom 92+\binom 66\binom 91=1485$$ But there is a simpler way; first pick 4 oranges from those 6 oranges (so that we have picked at least 4 oranges), and then choose 3 fruits from the remaining fruits (2 oranges, 5 apples and 4 bananas which sum to 11 fruits). This way we do have picked 7 fruits and at least 4 of them are oranges but the result is: $$\binom 64\binom {11}3=2475$$ which is different from the actual answer. How can I rigorously check which method works without having to list all the combinations?

• I am using the rule of sum and rule of product. The binomial coefficients represent the number of combinations. I want know what went wrong that I got two different answers from two different but seemingly logical and correct methods of counting. How can I avoid such counting errors by rigorously proving the right method without having to list all of the combinations? I don't know much beyond the rules I stated, so please consider that while answering. – M.Mahdi Aug 3 at 1:15

Your second method counts each selection that has $$5$$ oranges $$\binom54=5$$ times, and each selection that has $$6$$ oranges $$\binom64=15$$ times. In each case you count the selection once for each set of $$4$$ oranges contained in it: any one of those sets could be the set of $$4$$ that you preselected.
• @M.Mahdi: In general you need to ask yourself whether a method of counting could possibly count the same collection or arrangement of things more than once. Notice that your second approach singles out $4$ oranges for special treatment. This means that whenever we end up choosing more than $4$ oranges, we’re actually counting our oranges in two groups, the original $4$ and the one or two that we added later. But we can get the same $5$ or $6$ oranges with a different division of them into original $4$ and extras, and each of those divisions gets counted separately. This is something to ... – Brian M. Scott Aug 3 at 1:56
• ... watch for whenever you single out a specific subset of some collection whose elements should be treated identically. In this case we’re singling out $4$ oranges as the ones originally chosen, distinguishing them from any that we choose later, even though all that we care about is which oranges were selected. – Brian M. Scott Aug 3 at 1:59