Question
Find r if $^5P_{r} { = ^6}{P_{r - 1}}$
✓
Answer
$^5P_r { = ^6}{P_{r - 1}}$
$\therefore \frac{{5!}}{{(5 - r)!}} = \frac{{6!}}{{(6- r+1)!}}$
$ \Rightarrow \frac{{5!}}{{(5 - r)!}} = \frac{{6 \times 5!}}{{(7 - r)(6 - r)(5 - r)!}}$
$ \Rightarrow 1 = \frac{6}{{(7 - r)(6 - r)}}$$ \Rightarrow {r^2} - 13r + 42 = 6$
$ \Rightarrow {r^2} - 13r + 36 = 0$$ \Rightarrow {r^2} - 9r - 4r + 36 = 0$
$ \Rightarrow r(r - 9) - 4(r - 9) = 0$$ \Rightarrow (r - 9)(r - 4) = 0$
$ \Rightarrow $ r = 9 or r = 4
Now r = 9 is not possible because r > n.
Thus r = 4
$\therefore \frac{{5!}}{{(5 - r)!}} = \frac{{6!}}{{(6- r+1)!}}$
$ \Rightarrow \frac{{5!}}{{(5 - r)!}} = \frac{{6 \times 5!}}{{(7 - r)(6 - r)(5 - r)!}}$
$ \Rightarrow 1 = \frac{6}{{(7 - r)(6 - r)}}$$ \Rightarrow {r^2} - 13r + 42 = 6$
$ \Rightarrow {r^2} - 13r + 36 = 0$$ \Rightarrow {r^2} - 9r - 4r + 36 = 0$
$ \Rightarrow r(r - 9) - 4(r - 9) = 0$$ \Rightarrow (r - 9)(r - 4) = 0$
$ \Rightarrow $ r = 9 or r = 4
Now r = 9 is not possible because r > n.
Thus r = 4
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