262hmm.. thats right, but then the options given are way much more!!
10 Answers
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.
30Well 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
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
262i 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!