I have problem approaching this question.
Suppose we have to choose 26 odd, distinct number from the set [1, 97]
inclusive. What's the probability that there is a pair of chosen
integers with a sum of 100?
The sample space: So there are $\frac{97-1}{2} +1 = 49 $ odd numbers in this set. Thus there are $C(49, 26)$ ways to select the numbers. For each selection, there is $C(26,2)$ ways to select a pair. Thus, there are $C(49, 26)\cdot C(26,2)$ ways to select a pair.
However, I'm stuck here. This seems overcounting to me. For example, first attempt choosing numbers from position 1 to position 26, and second attempt choosing numbers from position 2 to position 27. Thus there will be duplicated pairs of numbers if using the product rule.
Since the numbers chosen have to be distinct, we start off from the middle numbers, which are 49 and 51, and go backwards to 3 and 97 (so 1 will not create a sum of 100 with any number). Thus there are $\frac{49-3}{2}+1=24$ pairs of numbers that will have a sum of 100.
I don't know how to proceed from here. The double selection confuses me a lot. Would love any thoughts on this! Thanks a lot!
Best Answer
I think this is a "trick" question, at least in the sense that it doesn't rely on the binomial formula or the fundamental counting principle.
You have 24 pairs that sum to 100: $$(3,97)$$ $$(5,95)$$ $$(7,93)$$ $$...$$ $$(49,51)$$
And the number $1$.
You might be able to see that there are $2^{24}$ ways to pick 25 numbers with no pairs adding to 100 by picking $1$, and then one from each ordered pair.
But when we increase the pool to 26 numbers, there must be a "match".