How many diff ways can you select problems from 8 total problems (4 of 10 point ones and 4 of 5 point ones to equal exactly 30 total points?

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  • Anonymous
    8 months ago

    using generating functions

    pari/gp calculator

    gp > (((1 + x*y^5)^4)*((1 + x*y^10)^4))

    %1 = y^60*x^8 + (4*y^55 + 4*y^50)*x^7 + (6*y^50 + 16*y^45 + 6*y^40)*x^6 + (4*y^45 + 24*y^40 + 24*y^35 + 4*y^30)*x^5 + (y^40 + 16*y^35 + 36*y^30 + 16*y^25 + y^20)*x^4 + (4*y^30 + 24*y^25 + 24*y^20 + 4*y^15)*x^3 + (6*y^20 + 16*y^15 + 6*y^10)*x^2 + (4*y^10 + 4*y^5)*x + 1

    4*y^30*x^5 (draw 5)

    36*y^30*x^4 (draw 4)

    4*y^30*x^3 (draw 3)

    44 total ways

    this is the same as having a deck of 8 cards where 4 are 5 point values and 4 are 10 point values and one draws without replacement.

    this method is super easy and way less prone to making errors and gives all the sums with draws instantly.

    the 44 is the number of combinations where the order does not matter

  • 8 months ago

    No. of ways to select 3 "10 point problems"

    = ₄C₃

    = 4

    No. of ways to select 2 "10 point problems" and 2 "5 point problem"

    = ₄C₂ × ₄C₂

    = 6 × 6

    = 36

    No. of ways to select 1 "10 point problem" and 4 "5 point problems"

    = ₄C₁ × ₄C₄

    = 4 × 1

    = 4

    Total no. of ways to select problems to equal exactly 30 points

    = 4 + 36 + 4

    = 44

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