30I can't find mistakes to my logic but my answer is coming way below the given options. Can someone point out where I am going wrong?
Attempt:
Since we need to form three-element subsets from the given set whose sum of elements is even, we can form such a set only and only if the elements are : even-even-even or even-odd-odd, where ordering is unimportant considering the fact that it is a set after all.
Case 1 (even-even-even): No. of ways : 5C3 = 10
Case 2 (even-odd-odd): No. of ways: 5C1.5C2 = 50
Total no. of ways = 50+10= 60 [7] Big problem! [2]
262what i think is if the sum is even, then it just depends upon the third no. and not the first two which we chose.
30
Ashish Kothari
·2011-06-17 01:46:52
Well your logic seems correct too. In that case, you are suggesting an answer of 500? And I still can't get where my solution is flawed [7] , and also the fact that in the answer of 500 we might be considering the ordering which doesn't matter in case of a set.
262
Aditya Bhutra
·2011-06-17 11:07:33
hmm.. thats right, but then the options given are way much more!!
30
Ashish Kothari
·2011-06-17 11:11:23
Exactly.. whats the answer given?
62The question has to be wrong because the number of 3 element sub sets here is 10C3
Which is equal to 10.9.8/6 = 120
so the ones where the sum is even cant be higher :P
1i agree with nishant bhaiya
so answer should be 60 ?
30
Ashish Kothari
·2011-06-17 22:55:09
My vote is with you kreyszig! :D :)
262
Aditya Bhutra
·2011-06-17 23:02:08
i think there is a misprint, it shouldnt be subsets, but only arrangements... so then how do we proceed
1any selection of 3 numbers can either be odd or even. (equal chances)
so, 10C3/2 = (10*9*8/6)/2=60
cheers!
