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Xelga [282]
3 years ago
5

The class has n students, and n is an even number. The students are forming teams to work on their homework. Each team has exact

ly 2 students and each student has to appear in exactly one team. How many distinct ways are there to form the teams for the class?
Mathematics
1 answer:
yuradex [85]3 years ago
8 0

Answer:

Distinct ways  in which they all can form the teams for the class = n(n-1)

Step-by-step explanation:

Total numbers of students = n (even number)

Number of student in each team = 2

Number of distinct ways are there to form the teams for the class:

P^{n}_{k}=\frac{n!}{(n-k)!}

where = n = number of elements  = n

k = number of elements choose = 2

P^{n}_{2}=\frac{n!}{(n-2)!}=\frac{n\times (n-1)\times (n-2)!}{(n-2)!}

=n\times (n-1)

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Then, the zeros of the given quadratic function are:

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The polynomial remainder theorem says that a polynomial <em>p(x)</em> leaves a remainder of <em>p(k)</em> when it's divided by <em>x</em> - <em>k</em>.

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<em />

We're also given that <em>y</em> + 2 is a factor, which means dividing <em>p(y)</em> by it leaves no remainder, and so

<em>p</em>(-2) = 2(-2)³ + <em>b</em>(-2)² - <em>c</em>(-2) + <em>d</em> = 0

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<em />

Solve the system of equations in boldface. You can eliminate <em>d</em> from the first 3 to first solve for <em>b</em> and <em>c</em>, then solve for <em>d</em> :

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