To find the sum of every combination of 6 numbers from the given set (350, 442, 365, 290, 440, 440, 310), we first need to determine how many combinations we have.

The formula for the number of combinations is given by:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

Where $n = 7$ (since there are 7 numbers) and $r = 6$ (since we are selecting 6 at a time).

$\binom{7}{6} = \frac{7!}{6!(7-6)!} = 7$

So, there are 7 combinations. Now, let's calculate the sum for each combination:

1. Combination 1 (without 350): $442 + 365 + 290 + 440 + 440 + 310 = 2287$

2. Combination 2 (without 442): $350 + 365 + 290 + 440 + 440 + 310 = 2195$

3. Combination 3 (without 365): $350 + 442 + 290 + 440 + 440 + 310 = 2312$

4. Combination 4 (without 290): $350 + 442 + 365 + 440 + 440 + 310 = 2347$

5. Combination 5 (without 440 (first)): $350 + 442 + 365 + 290 + 440 + 310 = 2197$

6. Combination 6 (without 440 (second)): $350 + 442 + 365 + 290 + 440 + 310 = 2197$

7. Combination 7 (without 310): $350 + 442 + 365 + 290 + 440 + 440 = 2327$

Thus, the sums for all combinations of 6 numbers are:

• 2287
• 2195
• 2312
• 2347
• 2197
• 2197
• 2327

Would you like further details or clarifications?

Here are five related questions:

1. How do you compute combinations for larger sets of numbers?
2. What is the difference between permutations and combinations?
3. Can you compute sums for different-sized combinations (e.g., 5 numbers)?
4. How would you approach this problem using a different method?
5. Can the same number be used multiple times in combinations?

Tip: When working with combinations, order doesn’t matter, but for permutations, order does!