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Sagot :
Sure, let’s work through the problem step-by-step. We are given the equation:
[tex]\[ P(n, 2) + 24 = P(2n, 2) \][/tex]
where [tex]\(P(n, k)\)[/tex] is the number of permutations of [tex]\(n\)[/tex] items taken [tex]\(k\)[/tex] at a time.
The formula for permutations [tex]\(P(n, k)\)[/tex] is:
[tex]\[ P(n, k) = \frac{n!}{(n-k)!} \][/tex]
For [tex]\(P(n, 2)\)[/tex], the formula becomes:
[tex]\[ P(n, 2) = \frac{n!}{(n-2)!} \][/tex]
Similarly, for [tex]\(P(2n, 2)\)[/tex], the formula is:
[tex]\[ P(2n, 2) = \frac{(2n)!}{(2n-2)!} \][/tex]
Let's substitute these into the given equation:
[tex]\[ \frac{n!}{(n-2)!} + 24 = \frac{(2n)!}{(2n-2)!} \][/tex]
We know that:
[tex]\[ \frac{n!}{(n-2)!} = n \times (n-1) \][/tex]
So, the equation becomes:
[tex]\[ n(n-1) + 24 = \frac{(2n)!}{(2n-2)!} \][/tex]
Next, we can simplify [tex]\(\frac{(2n)!}{(2n-2)!}\)[/tex] as:
[tex]\[ \frac{(2n)!}{(2n-2)!} = (2n)(2n-1) \][/tex]
Thus, the equation now is:
[tex]\[ n(n-1) + 24 = (2n)(2n-1) \][/tex]
We expand both sides:
[tex]\[ n^2 - n + 24 = 4n^2 - 2n \][/tex]
Rearrange all terms to one side of the equation to set it to zero:
[tex]\[ n^2 - n + 24 = 4n^2 - 2n \][/tex]
[tex]\[ n^2 - n + 24 - 4n^2 + 2n = 0 \][/tex]
Combine like terms:
[tex]\[ -3n^2 + n + 24 = 0 \][/tex]
Multiply through by -1 to make the quadratic easier to work with:
[tex]\[ 3n^2 - n - 24 = 0 \][/tex]
This is a standard quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], with [tex]\(a = 3\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -24\)[/tex]. We can solve this using the quadratic formula:
[tex]\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the formula:
[tex]\[ n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-24)}}{2(3)} \][/tex]
[tex]\[ n = \frac{1 \pm \sqrt{1 + 288}}{6} \][/tex]
[tex]\[ n = \frac{1 \pm \sqrt{289}}{6} \][/tex]
[tex]\[ n = \frac{1 \pm 17}{6} \][/tex]
This gives us two potential solutions:
[tex]\[ n = \frac{1 + 17}{6} = \frac{18}{6} = 3 \][/tex]
[tex]\[ n = \frac{1 - 17}{6} = \frac{-16}{6} = -\frac{8}{3} \][/tex]
Since [tex]\(n\)[/tex] must be positive, we discard [tex]\(-\frac{8}{3}\)[/tex] and keep:
[tex]\[ n = 3 \][/tex]
Thus, the positive value of [tex]\(n\)[/tex] is [tex]\(\boxed{3}\)[/tex].
[tex]\[ P(n, 2) + 24 = P(2n, 2) \][/tex]
where [tex]\(P(n, k)\)[/tex] is the number of permutations of [tex]\(n\)[/tex] items taken [tex]\(k\)[/tex] at a time.
The formula for permutations [tex]\(P(n, k)\)[/tex] is:
[tex]\[ P(n, k) = \frac{n!}{(n-k)!} \][/tex]
For [tex]\(P(n, 2)\)[/tex], the formula becomes:
[tex]\[ P(n, 2) = \frac{n!}{(n-2)!} \][/tex]
Similarly, for [tex]\(P(2n, 2)\)[/tex], the formula is:
[tex]\[ P(2n, 2) = \frac{(2n)!}{(2n-2)!} \][/tex]
Let's substitute these into the given equation:
[tex]\[ \frac{n!}{(n-2)!} + 24 = \frac{(2n)!}{(2n-2)!} \][/tex]
We know that:
[tex]\[ \frac{n!}{(n-2)!} = n \times (n-1) \][/tex]
So, the equation becomes:
[tex]\[ n(n-1) + 24 = \frac{(2n)!}{(2n-2)!} \][/tex]
Next, we can simplify [tex]\(\frac{(2n)!}{(2n-2)!}\)[/tex] as:
[tex]\[ \frac{(2n)!}{(2n-2)!} = (2n)(2n-1) \][/tex]
Thus, the equation now is:
[tex]\[ n(n-1) + 24 = (2n)(2n-1) \][/tex]
We expand both sides:
[tex]\[ n^2 - n + 24 = 4n^2 - 2n \][/tex]
Rearrange all terms to one side of the equation to set it to zero:
[tex]\[ n^2 - n + 24 = 4n^2 - 2n \][/tex]
[tex]\[ n^2 - n + 24 - 4n^2 + 2n = 0 \][/tex]
Combine like terms:
[tex]\[ -3n^2 + n + 24 = 0 \][/tex]
Multiply through by -1 to make the quadratic easier to work with:
[tex]\[ 3n^2 - n - 24 = 0 \][/tex]
This is a standard quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], with [tex]\(a = 3\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -24\)[/tex]. We can solve this using the quadratic formula:
[tex]\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the formula:
[tex]\[ n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-24)}}{2(3)} \][/tex]
[tex]\[ n = \frac{1 \pm \sqrt{1 + 288}}{6} \][/tex]
[tex]\[ n = \frac{1 \pm \sqrt{289}}{6} \][/tex]
[tex]\[ n = \frac{1 \pm 17}{6} \][/tex]
This gives us two potential solutions:
[tex]\[ n = \frac{1 + 17}{6} = \frac{18}{6} = 3 \][/tex]
[tex]\[ n = \frac{1 - 17}{6} = \frac{-16}{6} = -\frac{8}{3} \][/tex]
Since [tex]\(n\)[/tex] must be positive, we discard [tex]\(-\frac{8}{3}\)[/tex] and keep:
[tex]\[ n = 3 \][/tex]
Thus, the positive value of [tex]\(n\)[/tex] is [tex]\(\boxed{3}\)[/tex].
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