Factorial Equation is a Thing, So Let’s Solve One

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  • Опубліковано 29 вер 2024
  • In this video we solve a nice-looking equation that includes several factorials. Moreover, we will verify in a rigorous manner that the roots we obtain are all possible roots.
    #equations #factorial #FactorialEquation #operation
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КОМЕНТАРІ • 14

  • @djgiesz4914
    @djgiesz4914 8 місяців тому +1

    You can also see that the left hand side is 120 and the right hand side by definition is 10 choose n. You can try testing the values, 0 1 and 2 won't work, but 10 choose 3 will be 120, so 3 is a solution. However in general n choose k is the same as n choose n-k, therefore n=10-3=7 is the other solution.

  • @nasrullahhusnan2289
    @nasrullahhusnan2289 5 місяців тому +1

    5!=2×3×4×5
    =2×5×3×3×4×2/(3×2)
    =10×9×8/6
    =(10!)/[(7!)6]
    =(10!)/[(7!)(3!)] as 6=3!
    As 5!=[n!(10-n)! the n=3 or 7

    • @CornerstonesOfMath
      @CornerstonesOfMath  5 місяців тому

      You solved the problem while keeping 5! on the left hand side. Good job!

  • @hmkl6813
    @hmkl6813 9 місяців тому +2

    Bruh, i drew pascals triangle(at least tried to)

    • @CornerstonesOfMath
      @CornerstonesOfMath  9 місяців тому

      That strategy actually works in this case, since the right hand side is given in the form of binomial coefficient (combination). And it won't take ridiculously long, since you will find the answer in the 10th row (or 11th row, depending on where you start) of the triangle. For more general type of equations, the method in the video is more applicable.

  • @Taokyle
    @Taokyle 10 місяців тому +2

    Basically 10Cr=120, find r
    Nice explanation tho

  • @КатяРыбакова-ш2д
    @КатяРыбакова-ш2д 9 місяців тому +1

    Разделим на 5!. Получим дробь, равную 1. Чтобы сократилась 7, знаменатель должен содержать 7!. => n=7 или 10-n=7, n=3. Ответ: 3; 7.

    • @CornerstonesOfMath
      @CornerstonesOfMath  9 місяців тому

      Sound logic, although n!(10-n)! must have 7! doesn't immediately leads to n=7. It only tells us that the denominator has to contain something no less than 7!, for example, like 7!, 8!, 9!, or 10!. But again, it is pretty easy to find out that values other than 7 (or equivalently 3) doesn't work.

  • @mahxylim7983
    @mahxylim7983 11 місяців тому +2

    nice explanation

  • @m.h.6470
    @m.h.6470 10 місяців тому +1

    How do you get that a(n) decreases from n=1 to 5 and increases from n=5 to 9 by calculating a(n+1)/a(n) >/< 1?
    I mean, it is true, I just don't see a direct connection between the two.

    • @CornerstonesOfMath
      @CornerstonesOfMath  10 місяців тому +1

      If a(n+1)/a(n) < 1, then a(n+1) < a(n), meaning that the sequence decreases.
      If a(n+1)/a(n) > 1, then a(n+1) > a(n), meaning that the sequence increases.