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

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.

Answer 2

To select 3 or more numbers from the numbers 1 to 8 such that 3 and 5 should be there.
The remaining numbers are: 1, 2, 4, 6, 7, 8.
As we have to select 1 or more numbers from these, we cannot select both 7 and 8.

Case 1: Let us select 7. We need to remove 8 and from the remaining 4 numbers, we can select none, one, two, three or all.
The number of ways = 2424 = 16

Case 2: Let us select 8. We need to remove 7 and from the remaining 4 numbers, we can select none, one, two, three or all.
The number of ways = 2424 = 16

Case 3: Let us remove 7 and 8. From the remaining 4 numbers, we can select one, two, three or all.
The number of ways = 24−124−1 = 15

Thus, total number of ways = 16 + 16 + 15 = 47