Rahul@gurome wrote:

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?

• 4950
  5050
  2500
  2550
  5000

Thanks for the wonderul explanation

I got an answer 2525 ,, as i failed to taken into account ( 51,51) etc

So i did not know to choose between C or D

Is it better to guess lesser number than higher in such situations

Please help

Regards

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.

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?



4950

5050

2500

2550

5000

thebigkats wrote:thanks Rahul. Probably your way was faster. this is how got to this solution (although took me full 3 min

for no 1: 100 (total 1)
2: 100, 99 (total 2)
3: 100, 99, 98 (total 3)
...
50: 100, .. 52, 51 (total 50)

51: 100..52, 50 (but 51, 50 was already counted so leave it) TOTAL = 49
52: 100..53, 51, 50, 49 (but 49, 50 and 51 are already counted so leave them). TOTAL = 48
53: 100..54, 53, 52, 51, 50, 49, 48 (but 53..48 are already counted so leave them) TOTAL = 47
...
99: 100.., 98, 97..2 (but last ones except 100 are already counted so leave them) TOTAL = 1
100: all are already counted so TOTAL=0

So for first 50, total = n(n+1)/2 = 50(50+1)/2 = 25*51=1275
from 52..10, total = n(n+1)/2 = 49 (49+1)/2 = 25*49=1225

Total = 1275 + 1225 = 2500

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?



4950

5050

2500

2550

5000