Let's solve the equation \(_{30}P_r = 21 \times _{30}P_{r-1}\) for \(r\).
We know that the permutation formula is given by:
[tex]\[
_{n}P_r = \frac{n!}{(n-r)!}
\][/tex]
Substituting \(n = 30\), the equation becomes:
[tex]\[
_{30}P_r = \frac{30!}{(30-r)!}
\][/tex]
and
[tex]\[
_{30}P_{r-1} = \frac{30!}{(30-(r-1))!} = \frac{30!}{(31-r)!}
\][/tex]
Given the original equation \(_{30}P_r = 21 \times _{30}P_{r-1}\), we can substitute these values in:
[tex]\[
\frac{30!}{(30-r)!} = 21 \times \frac{30!}{(31-r)!}
\][/tex]
To simplify, we can cancel the \(30!\) from both sides, as it is a common factor:
[tex]\[
\frac{1}{(30-r)!} = 21 \times \frac{1}{(31-r)!}
\][/tex]
This simplifies to:
[tex]\[
\frac{1}{(30-r)!} = \frac{21}{(31-r)!}
\][/tex]
Next, we cross-multiply to clear the fractions:
[tex]\[
(31-r)! = 21 \times (30-r)!
\][/tex]
Recall that \((31-r)!\) can be expanded as:
[tex]\[
(31-r)! = (31-r) \times (30-r)!
\][/tex]
Thus, the equation becomes:
[tex]\[
(31-r) \times (30-r)! = 21 \times (30-r)!
\][/tex]
Since \((30-r)!\) is a common factor on both sides, we can cancel it out:
[tex]\[
31-r = 21
\][/tex]
Solving for \(r\):
[tex]\[
31 - r = 21
\][/tex]
[tex]\[
r = 31 - 21
\][/tex]
[tex]\[
r = 10
\][/tex]
Therefore, the value of [tex]\(r\)[/tex] is [tex]\(10\)[/tex].
