Calculating Possibilities for Winning Tickets | Combinations Problem

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SUMMARY

The discussion centers on calculating the number of combinations for winning tickets from a total of 10 tickets, of which 3 are winning. When purchasing 6 tickets, the goal is to determine the number of combinations that result in at least one winning ticket. The relevant formula used is C_n^k = n! / (k!(n-k)!), and the solution involves calculating C_1_0^6 and C_7^3 to find the correct number of combinations. The final answer is derived from the expression C_1_0^6 - C_7^3.

PREREQUISITES
  • Understanding of combinatorial mathematics
  • Familiarity with the combination formula C_n^k
  • Basic factorial calculations
  • Knowledge of probability concepts related to winning outcomes
NEXT STEPS
  • Study advanced combinatorial problems using C_n^k
  • Learn about probability distributions in lottery scenarios
  • Explore the concept of complementary counting in combinatorics
  • Investigate real-world applications of combinations in game theory
USEFUL FOR

Students studying combinatorics, educators teaching probability, and anyone interested in solving mathematical problems related to lottery ticket combinations.

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Homework Statement



From 10 tickets of luck, only 3 are winning. If we buy 6 tickets, how many possibilities we have at least to have 1 winning ticket in our hand?

Homework Equations



[tex]C_n^k=\frac{n!}{k!(n-k)!}[/tex]

The Attempt at a Solution



6 tickets from 10, we can choose on [itex]C_1_0^6[/itex]. There are 4 tickets left. [itex]C_7^3[/tex] is the tickets which are not winning. Is the right answer:[tex]C_1_0^6[/tex] - [tex]C_7^3[/tex]<br /> <br /> ?[/itex]
 
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Focus on the "at least" part.
 
Sorry, but I can't figure out
 

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