Question

Chapter - permutations and combinations

find R if 5Pr = 6Pr-1 ​

Answer 1

Given :

$\sf {}^{5} P_r = {}^{6} P_{r - 1}$

To find :

The value of r

Solution :

The number of arrangement that can be made by "n" disimilar things taken "r" at a time where 'r' is in between 0 and n [ 0 < r ≤ n] and where repetition is not allowed is $\sf ^nP_r$

where,

$\dashrightarrow \sf^{n} P_r = \dfrac{n!}{(n - r)!}$

According to the question,

$\dashrightarrow\sf^{5} P_r = ^{6} P_{r - 1}$

$\dashrightarrow \: \: \sf \dfrac{5!}{(5 - r)!} = \dfrac{6!}{ \{6 - (r - 1) \}!}$

$\dashrightarrow \: \:\sf \dfrac{5!}{(5 - r)!} = \dfrac{6 \times 5!}{(6 - r + 1)!}$

$\dashrightarrow \: \:\sf \dfrac{5!}{(5 - r)!} = \dfrac{6 \times 5!}{(7 - r)!}$

$\dashrightarrow \: \:\sf \dfrac{1}{(5 - r)!} = \dfrac{6}{(7 - r)(6 - r)(5 - r)!}$

$\dashrightarrow \: \:\sf \dfrac{1}{1} = \dfrac{6}{(7 - r)(6 - r)}$

$\dashrightarrow \: \:\sf (7 - r)(6 - r) = 6$

$\dashrightarrow \: \:\sf 42 - 7r - 6r + {r}^{2} = 6$

$\dashrightarrow \: \:\sf 42 - 13r + {r}^{2} = 6$

$\dashrightarrow \: \:\sf {r}^{2} - 13r + 36 = 0$

By, splitting the middle terms,

$\dashrightarrow \: \:\sf {r}^{2} - (9 + 4)r + 36 = 0$

$\dashrightarrow \: \:\sf {r}^{2} - 9r - 4r + 36 = 0$

$\dashrightarrow \: \:\sf r(r - 9) - 4(r - 9) = 0$

$\dashrightarrow \: \:\sf (r - 4) (r - 9) = 0$

$\dashrightarrow \: \:\sf r = 4 \: and \: 9$

Therefore, r = 4 because 0≤ r ≤ 5