P and C doubt

Question

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

Answer 1

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

Answer 2

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

Comments

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}.

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yes got it

pegion hole is applicable here