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

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.