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

Let's break down the problem step by step:

Out of the 8 given numbers, 3 and 5 always have to be included. That means we need to find the combinations from the remaining numbers: 1, 2, 4, 6, 7, and 8. 

First, we have 2 numbers (3 and 5) fixed. We need to choose at least one more number to have a group of three distinct numbers. But we cannot choose both 7 and 8 together. 

Let's break this down case by case: 

1. **Selecting one number out of the 6 remaining numbers**: \[ _6C_1 = 6 \] 

2. **Selecting two numbers out of the 6 remaining numbers**: \[ _6C_2 = 15 \] 

But, one combination will have both 7 and 8. So, actual combinations: 15 - 1 = 14 

3. **Selecting three numbers out of the 6 remaining numbers**: \[ _6C_3 = 20 \] 

But, we need to remove combinations having both 7 and 8. They are: 

(7, 8, 1), (7, 8, 2), (7, 8, 4), and (7, 8, 6). So, there are 4 such combinations. 

Actual combinations: 20 - 4 = 16 

4. **Selecting four numbers out of the 6 remaining numbers**: 

\[ _6C_4 = 15 \] 

But, from these combinations, those having both 7 and 8 are: 

(7, 8, 1, 2), (7, 8, 1, 4), (7, 8, 1, 6), (7, 8, 2, 4), (7, 8, 2, 6), and (7, 8, 4, 6). 

So, there are 6 such combinations, Actual combinations: 15 - 6 = 9 

5. **Selecting five numbers out of the 6 remaining numbers**: \[ _6C_5 = 6 \] 

Here, each combination will necessarily have both 7 and 8, so no combination is valid in this case. 

Now, summing up the combinations from all cases: 6 + 14 + 16 + 9 = 45 

Therefore, there are 45 groups of three or more distinct numbers that satisfy the given conditions.