The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is(assume each number can only occur once)
Started 3 months ago by Shashank in
Explanatory Answer
Given that the numbers 3 and 5 should be present in every subset and contain at least 3 numbers in it.
First, we need to find the subsets possible
{3, 5, 1, 2, 4, 6, 7, 8} except 3 and 5, remaining all numbers have two possible outcomes that either it is in the set or out of the set.
So, the number of possible subsets containing at least 3 numbers is 26-1 = 63
(In this 26, we have a possibility that nothing is present from {1, 2, 4, 6, 7, 8}. So, we should remove that possibility.)
From these 63, we need to remove the subsets which have {3, 5, 7, 8}
{3, 5, 7, 8, 1, 2, 4, 6} except 3, 5, 7, 8 remaining all numbers {1, 2, 4, 6} have two possible outcomes that either it is in the set or out of the set.
So, the number of sets possible is 24 = 16
Therefore, the answer would be 63 - 16 = 47.
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