Suppose 5 numbers are selected at random from the set of numbers {1, 2, 3, 4, 5, 6, 7, 8}. Then which of the following statements is necessarily true about the selected numbers ?
Split the set into the pairs that add up to 9: \(\{1,8\}, \{2,7\}, \{3,6\}, \{4,5\}\). Every one of the eight numbers lies in exactly one of these four pairs, and this partition settles two of the four statements.
(C) is necessarily true. Choosing 5 numbers means placing 5 objects into 4 pairs, so by the pigeonhole principle at least one pair must be chosen completely. That pair sums to 9, so there is always at least one such pair.
(B) is necessarily true. Each complete pair uses up 2 of the 5 chosen numbers, so 3 complete pairs would require 6 numbers. With only 5 available, no more than 2 complete pairs can occur, which is exactly what "at most two pairs" asserts.
(A) is not necessarily true. The set has four odd numbers, 1, 3, 5 and 7. Selecting all four along with a single even number, say \(\{1, 3, 5, 7, 2\}\), gives a selection with only one even number, so "at least two even" can fail.
(D) is not necessarily true. The set has four even numbers, 2, 4, 6 and 8. Selecting all four together with any odd number, say \(\{2, 4, 6, 8, 1\}\), gives four even numbers, which breaks the limit of three.
Hence the statements that are necessarily true are that there are at most two pairs each summing to 9, and that there is at least one pair summing to 9.
Match List I with List II
Let R 1= {(1, 1), (2, 2), (3, 3)} and R 2 = {(1, 1), (1, 2), (1, 3), (1, 4)}
List I | List II |
(A) R 1∪ R 2 | (I) {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 3)} |
(B) R 1- R 2 | (II) {1, 1} |
(C) R 1∩ R 2 | (III) {(1, 2), (1, 3), (1, 4)} |
(D) R 2- R 1 | (IV) {(2, 2), (3, 3)} |
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