1thats today brilliants question...... c u can choose an integer between 20 and 200 .......n den for each integer u will have 40 optons...but sumtimz the case will be repeated!!so now try for urself..:)...
Find the number of ways in which we can choose 2 distinct integers from 1 to 200 so that the difference between them is atmost 20.
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8 Answers
11by d way...did u do that column matching of circles...wernt part b and d wrong????
1rohit,i cant get it....giv an example of case repitation////
62There are 2 methods .. one is the grid method .. the other is the "gadha ghoda" counting method... (both are equivalent though)
Lets take the smaller number to be 1,
the larger one can be chosen in 20 ways.. from 2 to 21
as long as the smaller number is less than equal to 180, the larger one can be chosen in 20 ways...
then the number of ways keeps decreasing to 19, 18, 17, .... 1
so the total number of ways is
180x20+(19+18+17+.....1) = 3600+190 = 3790 ways
*(There could be a small mistake because I am assuming that "between" means including the end points)
1i just guessed this one n got it right.....
wht i idid was i choose any arbitary number between 20 and 180 n for each of these there could be 40 oder choiices..... but then there would be repitation so i ddecided to divide by 2 and got a answer which was near to this :P....
1hey i too guessed this one right...i mean not just guessed...but my ans was coming around this only 3600
and i did it like....we can choose the smaller number from anywhere between 1 to 180 and then each has 20 choices for a bigger number....
so you get 3600
but i got my mistake now
and sir, there is absolutely no mistake in your ans
thnx...