goyalsau wrote:There are 100 tokens numbered from 1 to 100. In how many ways can two tokens be drawn simultaneously so that their sum is more than 100?
Careful!
This question is full of tricks and traps!
One trick is there are two categories, n ≤ 50 and n > 50 (as goyalsau mentioned). And another one is the tokens are drawn
simultaneously! Thus (2, 99) and (99, 2) are basically same.
Now, let's proceed to solve the problem.
As mentioned earlier, let's divide the numbers in two categories.
Category 1: n ≤ 50
For 1 : 1 set -> (1, 100)
For 2 : 2 set -> (2, 99), (2, 100)
For 3 : 3 set -> (3, 98), (3, 99), (3, 100)
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For 50 : 50 set -> (50, 51), ... , (50, 99), (50, 100)
Total number of sets = 1 + 2 + 3 + ... + 50
Category 1: n > 50
For 51 : 50 set -> (50, 51), (51, 52), ... ,(51, 100) No (51, 51)
For 52 : 51 set -> (49, 52), (50, 52), ... ,(52, 100) No (52, 52)
For 53 : 52 set -> (48, 53), (49, 53), ... ,(53, 100) No (53, 53)
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For 100 : 99 set -> (1, 100), (2, 100) ... , (99, 100) No (100, 100)
Total number of sets = 50 + 51 + 52 + ... + 99
Therefore total number of sets = (1 + 2 + 3 + ... + 50) + (50 + 51 + 52 + ... + 99)
= (1 + 2 + 3 + ... + 99) + 50
= (99)*(99 + 1)/2 + 50
= 99*50 + 50
= 50*100
= 5000
Now for the second trap, all the set are counted twice.
Thus effective number of set = 5000/2 = 2500
The correct answer is C.
i got 5000 too
