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Further Assignment

Practice 1

  1. Calculation for C₁₂: In combinatorial mathematics, the notation C₁₂ typically denotes the number of combinations of 12 items taken 1 at a time. The formula for combinations is given by:

   Cnk=racn!k!(n−k)!C_{n}^{k} = rac{n!}{k!(n-k)!}
   For C₁₂:
   - Here, n = 12 and k = 1.

   Substituting the values:
   C121=rac12!1!(12−1)!=rac12!1!imes11!=rac12imes11!1imes11!=12C_{12}^{1} = rac{12!}{1!(12-1)!} = rac{12!}{1! imes 11!} = rac{12 imes 11!}{1 imes 11!} = 12
   Therefore, C₁₂ = 12.

  1. Calculation for C: Further context is needed to clarify what C represents in this scenario. Assuming C denotes another combinatorial expression, we would apply the relevant combinatorial formula based on the specific scenario presented.

  2. Calculation for P² - C²: Assuming P represents a specific variable, this expression can be evaluated once the values of P and C have been established. The formula demonstrates a difference of squares. Therefore, it can be factored as:
       P2−C2=(P+C)(P−C)P^{2} - C^{2} = (P + C)(P - C)
       Specific values for P and C will be necessary to execute this calculation technically.

Practice 2

  • Round Robin Chess Tournament: In a round robin chess tournament, every player competes against every other player a specific number of times. If there are n players in the tournament, each player plays with every other player once resulting in a total of inom{n}{2} games played, which equals:
      extGames=racn(n−1)2ext{Games} = rac{n(n-1)}{2}. In this case, we know that there are 78 rounds played, hence:

       racn(n−1)2=78rac{n(n-1)}{2} = 78
       To solve for n, multiply both sides by 2:
       n(n−1)=156n(n-1) = 156
       This expands to:
       n2−n−156=0n^{2} - n - 156 = 0
       We can solve this quadratic equation using the quadratic formula:
       n=rac−bext±ext√(b2−4ac)2an = rac{-b ext{±} ext{√}(b^{2} - 4ac)}{2a} ; here a=1, b=-1, c=-156
       - The discriminant (b2−4ac)(b^{2} - 4ac) is calculated as:
       (−1)2−4imes1imes(−156)=1+624=625(-1)^{2} - 4 imes 1 imes (-156) = 1 + 624 = 625
       - Thus:
       n=rac1ext±252n = rac{1 ext{±} 25}{2} leads to two possible solutions:
       n=rac262=13extandn=rac−242=−12n = rac{26}{2} = 13 ext{ and } n = rac{-24}{2} = -12
       Since the number of players cannot be negative, we have 13 participants in the round robin tournament.

Practice 3

  • Choices for Borrowing Textbooks: Joanna has 8 different kinds of math textbooks available in the library, and she is required to borrow 2 of them. The calculation of choices can also be done using the combinations formula:

   Cnk=racn!k!(n−k)!C_{n}^{k} = rac{n!}{k!(n-k)!}
   Here, n = 8 and k = 2:

   C82=rac8!2!(8−2)!=rac8!2!6!=rac8imes72imes1=28C_{8}^{2} = rac{8!}{2!(8-2)!} = rac{8!}{2!6!} = rac{8 imes 7}{2 imes 1} = 28
   Thus, Joanna has 28 different choices of borrowing textbooks.

Practice 4

  • Triangles from Heptagon Vertices: A regular heptagon has 7 vertices. To find the number of triangles that can be formed by selecting any 3 vertices from the 7, we again apply the combinations formula:

   Cnk=racn!k!(n−k)!C_{n}^{k} = rac{n!}{k!(n-k)!}
   Here, n = 7 and k = 3:

   C73=rac7!3!(7−3)!=rac7!3!4!=rac7imes6imes53imes2imes1=35C_{7}^{3} = rac{7!}{3!(7-3)!} = rac{7!}{3!4!} = rac{7 imes 6 imes 5}{3 imes 2 imes 1} = 35
   Therefore, there are 35 triangles that can be formed using the vertices of a regular heptagon.