Answer
See the explanation
Work Step by Step
To calculate the number of combinations for picking four numbers from 1 to 50 without replacement, you can use the combination formula:
\[
C(n, k) = \frac{{n!}}{{k! \times (n - k)!}}
\]
where:
- \(n\) is the total number of options (50 in this case)
- \(k\) is the number of selections (4 in this case)
- \(C(n, k)\) represents the number of combinations.
So, for picking four numbers from 1 to 50:
\[
C(50, 4) = \frac{{50!}}{{4! \times (50 - 4)!}} = \frac{{50!}}{{4! \times 46!}}=230,300\approx 64 \text{ hours}
\]
After calculating this, you can find out how many seconds it would take to buy one ticket for each combination, as it takes one second to buy each ticket.
To extend this to picking five numbers instead of four, you would simply calculate \(C(50, 5)\) using the same formula.
\[
C(50, 5) = \frac{{50!}}{{5! \times (50 - 5)!}} = \frac{{50!}}{{5! \times 45!}}=2,118,760\approx 588 \text{ hours}
\]
This problem relates to the material from this chapter because ties directly into:
1. Algorithmic Complexity
Shows how combinatorial explosion makes brute-force solutions impractical as problem size increases.
2. Polynomial vs Non-Polynomial Time
Buying every ticket is a non-polynomial time approach as the number of combinations grows exponentially with the number of picks.
3. Search and Optimization
Highlights the need for efficient algorithms or heuristics when exhaustive search is infeasible.
