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

We need to find groups of 3 or more numbers from {1, 2, 3, 4, 5, 6, 7, 8} with two key constraints:

Must always include 3 and 5

Cannot include both 7 and 8 together

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Since 3 and 5 are always included, we can think of them as already "locked in" to every group. This means we only need to choose additional numbers from the remaining set: {1, 2, 4, 6, 7, 8}.

The constraint "7 and 8 are never included together" means we need to be careful when both 7 and 8 appear in our selection.

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For each group size, we'll use this approach:

Count all possible ways to choose the remaining numbers

Subtract the "bad" cases where both 7 and 8 are chosen together

This is much easier than trying to count the "good" cases directly!

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Groups of size 3: Already have {3, 5}, need to choose 1 more from {1, 2, 4, 6, 7, 8}

Total ways: ${{}}^{6}C_{1} = 6$

Bad cases: Since we're only choosing 1 number, we can't choose both 7 and 8, so no bad cases to subtract

Valid groups: $6 - 0 = 6$

Groups of size 4: Already have {3, 5}, need to choose 2 more from {1, 2, 4, 6, 7, 8}

Total ways: ${{}}^{6}C_{2} = 15$

Bad cases: Choosing both 7 and 8 = 1 way

Valid groups: $15 - 1 = 14$

Groups of size 5: Already have {3, 5}, need to choose 3 more from {1, 2, 4, 6, 7, 8}

Total ways: ${{}}^{6}C_{3} = 20$

Bad cases: If we choose both 7 and 8, we need 1 more from {1, 2, 4, 6} = ${{}}^{4}C_{1} = 4$ ways

Valid groups: $20 - 4 = 16$

Groups of size 6: Already have {3, 5}, need to choose 4 more from {1, 2, 4, 6, 7, 8}

Total ways: ${{}}^{6}C_{4} = 15$

Bad cases: If we choose both 7 and 8, we need 2 more from {1, 2, 4, 6} = ${{}}^{4}C_{2} = 6$ ways

Valid groups: $15 - 6 = 9$

Groups of size 7: Already have {3, 5}, need to choose 5 more from {1, 2, 4, 6, 7, 8}

Total ways: ${{}}^{6}C_{5} = 6$

Bad cases: If we choose both 7 and 8, we need 3 more from {1, 2, 4, 6} = ${{}}^{4}C_{3} = 4$ ways

Valid groups: $6 - 4 = 2$

Groups of size 8: Already have {3, 5}, need to choose all 6 from {1, 2, 4, 6, 7, 8}

This forces us to choose both 7 and 8, which violates our constraint

Valid groups: $0$

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Total number of valid groups = $6 + 14 + 16 + 9 + 2 = 47$

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When you have a "never together" constraint, complementary counting (Total - Bad cases) is often much simpler than direct counting. This technique is especially powerful in combination problems where certain elements have restrictions.