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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

Updated On: Jan 15, 2026
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Correct Answer: 47

Solution and Explanation

Given the constraints that every subset must include the numbers 3 and 5, and each subset must contain a minimum of 3 elements, we proceed with a step-by-step solution.

The initial set is: \(\{3, 5, 1, 2, 4, 6, 7, 8\}\). As 3 and 5 are mandatory inclusions, we focus on the remaining elements: \(\{1, 2, 4, 6, 7, 8\}\).

Step 1: Determine the Total Number of Possible Subsets

For each of the remaining 6 numbers, there are two possibilities: inclusion or exclusion. This yields a total of \( 2^6 = 64 \) potential subsets.

We must exclude the case where none of the remaining numbers are selected, which corresponds to the empty set. Therefore, the total number of valid subsets meeting the initial criteria is:

\[ 64 - 1 = 63 \]

Step 2: Identify Subsets Containing Both 7 and 8

Next, we identify subsets that include both 7 and 8. With 3, 5, 7, and 8 fixed, the remaining numbers available for selection are \(\{1, 2, 4, 6\}\).

Each of these 4 numbers can either be included or excluded, resulting in \( 2^4 = 16 \) subsets that contain both 7 and 8.

Step 3: Final Calculation

To find the number of subsets that include 3 and 5, have at least 3 elements, but do not contain both 7 and 8 simultaneously, we subtract the count of invalid subsets from the total valid subsets:

\[ 63 - 16 = 47 \]

Conclusion

Consequently, there are 47 subsets that satisfy all conditions: containing at least 3 numbers, always including 3 and 5, and not including both 7 and 8 together.

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