Correct Answer: Option C

Explanation

To solve the equation \(^{n}P_{3} - 6(^{n}C_{4}) = 0\), we start by using the formulas for permutations and combinations:

\(^{n}P_{r} = \frac{n!}{(n-r)!}\)
\(^{n}C_{r} = \frac{n!}{r!(n-r)!}\)

\(^{n}P_{3} = \frac{n!}{(n-3)!}\)

\(^{n}C_{4} = \frac{n!}{4!(n-4)!}\)

\(\frac{n!}{(n-3)!} - 6 \left(\frac{n!}{4!(n-4)!}\right) = 0\)

Factor out \(n!\)

Factoring \(n!\) from both terms gives:

\(n! \left(\frac{1}{(n-3)!} - \frac{6}{4!(n-4)!}\right) = 0\)

Since \(n! \neq 0\), we can simplify to:

\(\frac{1}{(n-3)!} - \frac{6}{4!(n-4)!} = 0\)

Recall that \(4! = 24\), so:

\(\frac{6}{4!(n-4)!} = \frac{6}{24(n-4)!} = \frac{1}{4(n-4)!}\)

The equation now is:

\(\frac{1}{(n-3)!} - \frac{1}{4(n-4)!} = 0\)

Multiply through by \(4(n-3)!(n-4)!\) gives:

\(4(n-4)! - (n-3)! = 0\)

Recall that \((n-3)! = (n-3)(n-4)!\), so:

\(4(n-4)! - (n-3)(n-4)! = 0\)

\((n-4)!\)

Factoring out \((n-4)!\) gives:

\((n-4)! \left(4 - (n-3)\right) = 0\)

Since \((n-4)! \neq 0\), we have:

\(4 - (n-3) = 0\)

Solving for \(n\):

\(n - 3 = 4 \implies n = 7\)

The value of n is: 7

There is an explanation video available below.

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