Andrew selects 6 distinct numbers from 1—49 for a lottery ticket. How many different tickets are possible (order does not matter)?
There are 13,983,816 different lottery tickets possible.
In a lottery where order does not matter, the number of combinations can be calculated using the binomial coefficient formula, specifically C(n, k), which represents the number of ways to choose k elements from a set of n elements. For this case, C(49, 6) yields 13,983,816 distinct combinations.
This number would be the result of calculating combinations for a different scenario, such as selecting 5 numbers from 20. It does not apply to the given problem, which requires selecting 6 numbers from a total of 49.
This option could represent a miscalculation or a different selection of numbers, perhaps by mistakenly considering the arrangement of numbers (permutations) rather than combinations. However, it does not correspond to the correct method of calculating C(49, 6).
This is the correct answer, calculated using the binomial coefficient C(49, 6), which equals 49! / (6! * (49-6)!). This formula accounts for the fact that the order of the numbers does not matter, providing the total number of unique lottery ticket combinations possible.
This choice appears arbitrary and does not reflect a logical calculation based on the principles of combinations. It significantly underestimates the number of possible combinations when selecting 6 numbers from a pool of 49.
In summary, the calculation of lottery ticket combinations, where order does not matter, leads us to the conclusion that 13,983,816 unique tickets can be formed from 6 distinct selections out of 49 numbers. Understanding the application of combinations versus permutations is crucial in reaching the correct answer and recognizing the vast number of possibilities in this lottery scenario.
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