Question

if P (5,r) = 2 X P(6, r-1) find r

Answer 1

Hi...☺

Here is your answer....✌

GIVEN THAT,

${} ^{5}P_{r} = 2 \times {}^{6}P_{r - 1} \\ \\ = > \frac{5!}{(5 - r)!} = 2 \times \frac{6!}{(6 - (r - 1))!} \\ \\ \frac{5!}{(5 - r)!} = 2 \times \frac{6\times 5!}{(6 - r + 1)!} \\ \\ \frac{1}{(5 - r)!} = 2 \times \frac{6}{(7 - r)!} \\ \\ \frac{1}{(5 - r)!} = \frac{12}{(7 - r) \times (6 - r) \times (5 - r)!} \\ \\ 1 = \frac{12}{(7 - r) \times (6 - r) } \\ \\ (7 - r)(6 - r) = 12 \\ \\ 42 - 13r + r {}^{2} = 12 \\ \\ {r}^{2} - 13r + 30 = 0 \\ \\ {r}^{2} - 10r - 3r + 30 = 0 \\ \\ {r}(r - 10) - 3(r - 10) = 0 \\ \\ (r - 3)(r - 10) = 0 \\ \\ = > r = 3 \: \: or \: \: r = 10 \\ \\ => r = 3$

Since,
Value of r can't be greater than 5

[ As we know r ≤ n ]

HENCE,
r = 3

Answer 2

AnswEr:

We have,

$\qquad \sf \: P(5,r) = 2.P(6,r - 1) \\ \\ \hookrightarrow \sf \: \frac{5!}{(5 - r)!} = 2. \frac{6!}{(6 - (r - 1))!} \\ \\ \hookrightarrow \sf \frac{5!}{(5 - r)!} = \frac{2 \times 6 \times 5!}{(7 - r)!} \\ \\ \hookrightarrow \sf \frac{5!}{(5 - r)!} = \frac{12 \times 5!}{(7 - r)(6 - r)(5 - r)!} \\ \\ \hookrightarrow \sf1 = \frac{12}{(7 - r)(6 - r)} \\ \\ \hookrightarrow \sf \: (7 - r)(6 - r) = 12 \\ \\ \hookrightarrow \sf \: (7 - 2)(6 - r) = 4 \times 3 \\ \\ \hookrightarrow \sf \: 7 - r = 4 \\ \\ \hookrightarrow \sf \: r = 3 \\ \\ \therefore \sf \: r = 3$