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

Select 3 or more numbers from the set \(\{1, 2, 3, 4, 5, 6, 7, 8\}\) such that numbers 3 and 5 are always included. The additional restriction is that you cannot select both 7 and 8 together.

Initial Setup 

Since 3 and 5 must be included, the remaining numbers to consider for selection are: \[ \{1, 2, 4, 6, 7, 8\}. \] You must select at least one number from these remaining six.

Restriction

7 and 8 cannot both be selected simultaneously.

Cases to Consider

Case 1: Select 7 (exclude 8)

Numbers to select additionally: \(\{1, 2, 4, 6\}\) along with 7.
From these 4 numbers, you can select anywhere from 0 to 4.
Number of ways: \[ \sum_{k=0}^4 \binom{4}{k} = 2^4 = 16. \]

Case 2: Select 8 (exclude 7)

Similarly, choose from \(\{1, 2, 4, 6\}\) and 8.
Number of ways also: \[ 2^4 = 16. \]

Case 3: Exclude both 7 and 8

Now choose from \(\{1, 2, 4, 6\}\) only, selecting at least one since 3 and 5 are already included.
Number of ways: \[ \sum_{k=1}^4 \binom{4}{k} = 2^4 - \binom{4}{0} = 16 - 1 = 15. \]

Total Number of Ways

\[ 16 + 16 + 15 = 47 \]

✅ Final Answer: 47 ways