Question

The number of groups of three or more distinct numbers that can be chosen from1, 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

Answer 1

Correct Answer: 47

Given that the numbers 3 and 5 should be present in every subset and each subset should contain at least 3 numbers, let’s solve the problem step by step:

We start with the set: \(\{3, 5, 1, 2, 4, 6, 7, 8\}\). Since 3 and 5 must be included in every subset, we focus on the remaining numbers: \(\{1, 2, 4, 6, 7, 8\}\).

Step 1: Number of Possible Subsets

For each of the remaining numbers \(\{1, 2, 4, 6, 7, 8\}\), there are two possible outcomes: either the number is in the set or it is out of the set. Therefore, the total number of subsets is:

\[ 2^6 = 64 \]

Now, we need to subtract the case where no number from \(\{1, 2, 4, 6, 7, 8\}\) is included. This represents the empty subset. So, the total number of valid subsets is:

\[ 64 - 1 = 63 \]

Step 2: Subsets Containing Both 7 and 8

Now, we need to exclude the subsets that have both 7 and 8 in them. In this case, we fix 3, 5, 7, and 8 in the subset. The remaining numbers to choose from are \(\{1, 2, 4, 6\}\).

Each of the remaining numbers \(\{1, 2, 4, 6\}\) has two possibilities: either it is included or not. Thus, the number of subsets where both 7 and 8 are included is:

\[ 2^4 = 16 \]

Step 3: Final Calculation

Now, we subtract the invalid subsets (those containing both 7 and 8) from the total number of valid subsets:

\[ 63 - 16 = 47 \]

Conclusion

Therefore, the number of subsets containing at least 3 numbers, including 3 and 5, but not both 7 and 8 together, is 47.