5 votes 5 votes How many number must be chosen from site {1 2 3 4 5 6 7 8 }such that at least two of them must have sum equal to 9? A.28 B.9 C.5 D.10 Combinatory pigeonhole-principle discrete-mathematics + – Surya Dhanraj asked Oct 10, 2017 • retagged Nov 26, 2019 by KUSHAGRA गुप्ता Surya Dhanraj 470 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 9 votes 9 votes With 2 elements pairs which give sum as 9 $= \{(1,8),(2,7),(3,6),(4,5)\}$ So choosing 1 elements from each group = 4 elements (in worst case 4 elements will be $\{1,2,3,4\}$ or $\{ 8,7,6,5\}$) Now using pigeonhole principle = we need to choose 1 more element so that sum will definitely be 9 so Number of elements = 4 +1 = 5 Prashant. answered Oct 11, 2017 • selected Oct 11, 2017 by Arjun Prashant. comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes With (1,2,3,4) any two element cannot create sum=9 So, now taking 5 in that group (1,2,3,4,5) we get sum of (4,5) which will give sum of 9 So, atmost 5 number we must have to choose to get 9 srestha answered Oct 11, 2017 • edited Oct 11, 2017 by srestha srestha comment Share Follow See all 2 Comments See all 2 2 Comments reply Arjun commented Oct 11, 2017 reply Follow Share The question is for selecting numbers and the condition is that at least "two of them" must sum to 9. i.e., {5, 4, 2}, {5, 4, 2, 6} etc. are valid but not {1, 2, 6}. 1 votes 1 votes srestha commented Oct 11, 2017 reply Follow Share yes got it pegion hole is applicable here 0 votes 0 votes Please log in or register to add a comment.