when i work over combination problems , i run into two questions.Both of them were talking about same concept ,but their solutions were different.
Fist question:
A student write these letters to whiteboard a,b,b,c,c,c,d,d,d,d. After that, his teacher wants to erase some letters in order that there left only four letters which are distinct from each other.How many ways are there to do it?
Solution: $\binom21\times\binom32\times\binom43=24$ ways .We chose one b from two bs and two c from three cs and three d from four ds.
Second question:
Suppose a shipment of 100 computers contains four defective computers. We choose a sample of six computers. How many different samples are there if the defective computers are distinct from each other and the unfaulty ones are identical?
Solution: no defective computer = 1 since all unfaulty computers are identical
When we choose same objects in the former question , we used combination.For example , we selected 3 ds among 4 ds , and all of ds were same.On the other hand, in the second question, we said that all of them are the same of each other.Because of this, there is only one way to select.
My question is that which one is true solution.Are not both of them same question type.Why didnt we say that all of ds are same and there is only one selection in the first.Otherwise, why didnt we we make use of $\binom{94}{6}$ to select 6 identical computer in the second question?
No, the questions are not of the same type. In the first question all ten letters are distinct. For instance, erasing the first $d$ on the whiteboard is different from erasing the third $d$, and we can see that it is different. In the other problem we are asked to assume that the $94$ computers that are not defective really are indistinguishable, so that we cannot tell one set of $6$ of them from another. This is not an entirely realistic assumption, but we’re asked to make it for the sake of the problem.