Solved on Jan 23, 2024

Solve for $x$ in the equation ${ }_{n} P_{4}=4{ }_{n-1} P_{3}$ , where ${ }_{n} P_{k}$ represents the number of permutations of $k$ items from a set of $n$ items.

STEP 1

Assumptions
1. ${ }_{n} P_{4}$ and $4{ }_{n-1} P_{3}$ are permutations.
2. The notation ${ }_{n} P_{k}$ stands for the number of permutations of $n$ distinct objects taken $k$ at a time.
3. The formula for a permutation is given by ${ }_{n} P_{k} = \frac{n!}{(n-k)!}$ .
4. We need to solve for $x$ in the equation ${ }_{n} P_{4}=4{ }_{n-1} P_{3}$ .

STEP 2

Write down the permutation formula for ${ }_{n} P_{4}$ .
${ }_{n} P_{4} = \frac{n!}{(n-4)!}$

STEP 3

Write down the permutation formula for ${ }_{n-1} P_{3}$ .
${ }_{n-1} P_{3} = \frac{(n-1)!}{(n-1-3)!}$

STEP 4

Substitute the permutation formulas into the given equation.
$\frac{n!}{(n-4)!} = 4 \times \frac{(n-1)!}{(n-4)!}$

STEP 5

Since the denominators are the same, we can equate the numerators.
$n! = 4 \times (n-1)!$

STEP 6

Expand the factorial on the right side of the equation.
$n! = 4 \times (n-1) \times (n-2) \times (n-3) \times (n-4)!$

STEP 7

Since $(n-4)!$ is a common factor on both sides, we can cancel it out.
$n = 4 \times (n-1) \times (n-2) \times (n-3)$

STEP 8

Expand the right side of the equation.
$n = 4(n^3 - 6n^2 + 11n - 6)$

STEP 9

Distribute the 4 across the terms on the right side.
$n = 4n^3 - 24n^2 + 44n - 24$

STEP 10

Rearrange the equation to bring all terms to one side and set the equation to zero.
$4n^3 - 24n^2 + 44n - 24 - n = 0$

STEP 11

Combine like terms.
$4n^3 - 24n^2 + 43n - 24 = 0$

STEP 12

Now we have a cubic equation to solve for $n$ . We look for rational roots using the Rational Root Theorem, which suggests that any rational solution of the equation is a factor of the constant term divided by a factor of the leading coefficient.
The factors of the constant term (-24) are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24$ , and the factors of the leading coefficient (4) are $\pm1, \pm2, \pm4$ .

STEP 13

Test the possible rational roots by substituting them into the equation until we find a root.
After testing the possible roots, we find that $n = 4$ is a root of the equation.

STEP 14

Use synthetic division or polynomial division to divide the cubic polynomial by $(n - 4)$ to find the other factors.

STEP 15

After performing the division, we get a quadratic equation.

STEP 16

Solve the quadratic equation using the quadratic formula or by factoring if possible.

STEP 17

The solutions of the quadratic equation will give us the remaining possible values for $n$ .

STEP 18

Since $n$ must be a positive integer greater than or equal to 4 (because we cannot take a permutation of a negative number of objects, and $n$ must be at least 4 for ${ }_{n} P_{4}$ to make sense), we discard any negative or non-integer solutions.

STEP 19

The remaining positive integer solution(s) for $n$ are the answer(s) to the original problem.
The solution for $n$ is $n = 4$ .

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Solve for $x$ in the equation ${ }_{n} P_{4}=4{ }_{n-1} P_{3}$ , where ${ }_{n} P_{k}$ represents the number of permutations of $k$ items from a set of $n$ items.

STEP 1

STEP 2

Write down the permutation formula for ${ }_{n} P_{4}$ .
${ }_{n} P_{4} = \frac{n!}{(n-4)!}$

STEP 3

Write down the permutation formula for ${ }_{n-1} P_{3}$ .
${ }_{n-1} P_{3} = \frac{(n-1)!}{(n-1-3)!}$

STEP 4

Substitute the permutation formulas into the given equation.
$\frac{n!}{(n-4)!} = 4 \times \frac{(n-1)!}{(n-4)!}$

STEP 5

Since the denominators are the same, we can equate the numerators.
$n! = 4 \times (n-1)!$

STEP 6

Expand the factorial on the right side of the equation.
$n! = 4 \times (n-1) \times (n-2) \times (n-3) \times (n-4)!$

STEP 7

Since $(n-4)!$ is a common factor on both sides, we can cancel it out.
$n = 4 \times (n-1) \times (n-2) \times (n-3)$

STEP 8

Expand the right side of the equation.
$n = 4(n^3 - 6n^2 + 11n - 6)$

STEP 9

Distribute the 4 across the terms on the right side.
$n = 4n^3 - 24n^2 + 44n - 24$

STEP 10

Rearrange the equation to bring all terms to one side and set the equation to zero.
$4n^3 - 24n^2 + 44n - 24 - n = 0$

STEP 11

Combine like terms.
$4n^3 - 24n^2 + 43n - 24 = 0$

STEP 12

STEP 13

Test the possible rational roots by substituting them into the equation until we find a root.
After testing the possible roots, we find that $n = 4$ is a root of the equation.

STEP 14

Use synthetic division or polynomial division to divide the cubic polynomial by $(n - 4)$ to find the other factors.

STEP 15

After performing the division, we get a quadratic equation.

STEP 16

Solve the quadratic equation using the quadratic formula or by factoring if possible.

STEP 17

The solutions of the quadratic equation will give us the remaining possible values for $n$ .

STEP 18

Since $n$ must be a positive integer greater than or equal to 4 (because we cannot take a permutation of a negative number of objects, and $n$ must be at least 4 for ${ }_{n} P_{4}$ to make sense), we discard any negative or non-integer solutions.

STEP 19

The remaining positive integer solution(s) for $n$ are the answer(s) to the original problem.
The solution for $n$ is $n = 4$ .

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Solve for xxx in the equation nP4=4n−1P3{ }_{n} P_{4}=4{ }_{n-1} P_{3}n​P4​=4n−1​P3​, where nPk{ }_{n} P_{k}n​Pk​ represents the number of permutations of kkk items from a set of nnn items.

STEP 1

STEP 2

Write down the permutation formula for nP4{ }_{n} P_{4}n​P4​.nP4=n!(n−4)!{ }_{n} P_{4} = \frac{n!}{(n-4)!}n​P4​=(n−4)!n!​

STEP 3

Write down the permutation formula for n−1P3{ }_{n-1} P_{3}n−1​P3​.n−1P3=(n−1)!(n−1−3)!{ }_{n-1} P_{3} = \frac{(n-1)!}{(n-1-3)!}n−1​P3​=(n−1−3)!(n−1)!​

STEP 4

Substitute the permutation formulas into the given equation.n!(n−4)!=4×(n−1)!(n−4)!\frac{n!}{(n-4)!} = 4 \times \frac{(n-1)!}{(n-4)!}(n−4)!n!​=4×(n−4)!(n−1)!​

STEP 5

Since the denominators are the same, we can equate the numerators.n!=4×(n−1)!n! = 4 \times (n-1)!n!=4×(n−1)!

STEP 6

Expand the factorial on the right side of the equation.n!=4×(n−1)×(n−2)×(n−3)×(n−4)!n! = 4 \times (n-1) \times (n-2) \times (n-3) \times (n-4)!n!=4×(n−1)×(n−2)×(n−3)×(n−4)!

STEP 7

Since (n−4)!(n-4)!(n−4)! is a common factor on both sides, we can cancel it out.n=4×(n−1)×(n−2)×(n−3)n = 4 \times (n-1) \times (n-2) \times (n-3)n=4×(n−1)×(n−2)×(n−3)

STEP 8

Expand the right side of the equation.n=4(n3−6n2+11n−6)n = 4(n^3 - 6n^2 + 11n - 6)n=4(n3−6n2+11n−6)

STEP 9

Distribute the 4 across the terms on the right side.n=4n3−24n2+44n−24n = 4n^3 - 24n^2 + 44n - 24n=4n3−24n2+44n−24

STEP 10

Rearrange the equation to bring all terms to one side and set the equation to zero.4n3−24n2+44n−24−n=04n^3 - 24n^2 + 44n - 24 - n = 04n3−24n2+44n−24−n=0

STEP 11

Combine like terms.4n3−24n2+43n−24=04n^3 - 24n^2 + 43n - 24 = 04n3−24n2+43n−24=0

STEP 12

STEP 13

Test the possible rational roots by substituting them into the equation until we find a root.After testing the possible roots, we find that n=4n = 4n=4 is a root of the equation.

STEP 14

Use synthetic division or polynomial division to divide the cubic polynomial by (n−4)(n - 4)(n−4) to find the other factors.

STEP 15

After performing the division, we get a quadratic equation.

STEP 16

Solve the quadratic equation using the quadratic formula or by factoring if possible.

STEP 17

The solutions of the quadratic equation will give us the remaining possible values for nnn.

STEP 18

Since nnn must be a positive integer greater than or equal to 4 (because we cannot take a permutation of a negative number of objects, and nnn must be at least 4 for nP4{ }_{n} P_{4}n​P4​ to make sense), we discard any negative or non-integer solutions.

STEP 19

The remaining positive integer solution(s) for nnn are the answer(s) to the original problem.The solution for nnn is n=4n = 4n=4.

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Solve for $x$ in the equation ${ }_{n} P_{4}=4{ }_{n-1} P_{3}$ , where ${ }_{n} P_{k}$ represents the number of permutations of $k$ items from a set of $n$ items.

Write down the permutation formula for ${ }_{n} P_{4}$ .
${ }_{n} P_{4} = \frac{n!}{(n-4)!}$

Write down the permutation formula for ${ }_{n-1} P_{3}$ .
${ }_{n-1} P_{3} = \frac{(n-1)!}{(n-1-3)!}$

Substitute the permutation formulas into the given equation.
$\frac{n!}{(n-4)!} = 4 \times \frac{(n-1)!}{(n-4)!}$

Since the denominators are the same, we can equate the numerators.
$n! = 4 \times (n-1)!$

Expand the factorial on the right side of the equation.
$n! = 4 \times (n-1) \times (n-2) \times (n-3) \times (n-4)!$

Since $(n-4)!$ is a common factor on both sides, we can cancel it out.
$n = 4 \times (n-1) \times (n-2) \times (n-3)$

Expand the right side of the equation.
$n = 4(n^3 - 6n^2 + 11n - 6)$

Distribute the 4 across the terms on the right side.
$n = 4n^3 - 24n^2 + 44n - 24$

Rearrange the equation to bring all terms to one side and set the equation to zero.
$4n^3 - 24n^2 + 44n - 24 - n = 0$

Combine like terms.
$4n^3 - 24n^2 + 43n - 24 = 0$

Test the possible rational roots by substituting them into the equation until we find a root.
After testing the possible roots, we find that $n = 4$ is a root of the equation.

Use synthetic division or polynomial division to divide the cubic polynomial by $(n - 4)$ to find the other factors.

The solutions of the quadratic equation will give us the remaining possible values for $n$ .

Since $n$ must be a positive integer greater than or equal to 4 (because we cannot take a permutation of a negative number of objects, and $n$ must be at least 4 for ${ }_{n} P_{4}$ to make sense), we discard any negative or non-integer solutions.

The remaining positive integer solution(s) for $n$ are the answer(s) to the original problem.
The solution for $n$ is $n = 4$ .