13no options ...given
it came in ISI today..... i gooffed up in the middle...:P
3 numbers are chosen from a set of {1,2,3...4n}...the number of ways of choosing such that the sum of them is divisible by 4..??
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11 Answers
21OPTIONS wud it make it quite easier by putting n =1......
even otherwise is possible.... 'ABHI'([3]) karta huun
13
62hint:
out of these,
4k
4k+1
4k+2
4k+3
all 3 are div by 4.. nC3
take other cases
13:(...bhaaiya..i din get u...:(
i m getting dull ...day by day...:(
1luk at the set of numbers..
1 2 3 4
5 6 7 8
..........
..........
general terms wud be like,
4K 4K+1 4K+2 4K+3
we have n such sets...
the number of ways in which we can get 4, using 0,1,2,3 are
case1) 0+1+3
case2) 0+2+2
case3)1+2+1
case4)0+0+0 (multiple of 4)
for case 1) we need 3 numbers such like 4K, 4K+1,4K+3
" 2) 4K,4K+2,4K+2
" 3) 4K+1,4K+2,4K+1
" 4) 4K,4K,4K
so accordingly we have
1) n*n*n=n3
2)n*nC2
3)n*nC2
4)nC3
so the total number of ways is,
N=n3+2n*nC2+nC3
ab iska totalling kariyo :)
(i mite hav missed out some cases, but the method is quite the same)
cheers!!!
17,3,2 is of form 4K+3, 4K+3,4K+2
toatl 3+3+2 = 8
possible .. gordo missed it
13So
N=n3+3n*nC2+nC3
seems to be the ans....
i missed a few cases in the exm...[17]
1yes .. better cry right now .. rather than crying out when the results come ... :(
1ohh sorrrryyy( arghh!!)...yeah missed out dat 2,3,3 walla case...
anyways, i think dat shud be it...wat say?
cheers!!