Question

5 4Pr = 2 5Pr (statistics)​

Answer 1

Given : $5 . ^4 P_r = 2 . ^5 P_r$

To Find : Value of r

Solution:

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

$5 . ^4 P_r = 2 . ^5 P_r$

=> 5 . 4!/(4 - r)!   = 2.  5!/(5 - r)!

=> 5. 4 * 3 * . . . (4 - r + 1)   = 2.  5* 4 * . . .  (5 - r + 1)

=>  5. 4 * 3 * . . . (5 - r + 1) (4 - r + 1)   = 2.  5* 4 * . . . (5 - r + 1)

5. 4 * 3 * . . . (5 - r + 1)  gets cancelled out

=   (4 - r + 1) = 2.  

=> 5 - r = 2

=> r = 3

Verification :

$5 . ^4 P_3 = 2 . ^5 P_3$

$5. ^4 P_4 =5. \dfrac{4!}{(4-3)!}=5!$

$2. ^5 P_4 =2. \dfrac{5!}{(5-3)!}=2. \dfrac{5!}{2!} = 5!$

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Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{ 5. \: {}^{4}P_r = 2. \: {}^{5}P_r}$

TO DETERMINE

The value of r

EVALUATION

$\displaystyle \sf{ 5. \: {}^{4}P_r = 2. \: {}^{5}P_r}$

$\displaystyle \sf{ \implies \: 5. \frac{4!}{(4 - r)!} \: = 2. \: \frac{5!}{(5 - r)!} }$

$\displaystyle \sf{ \implies \: 5. \frac{4!}{(4 - r)!} \: = 2. \: \frac{5 \times 4!}{(5 - r) \times (4 - r)!} }$

$\displaystyle \sf{ \implies \: 5 \: = 2. \: \frac{5 }{(5 - r) } }$

$\displaystyle \sf{ \implies \: \frac{2 }{(5 - r) } = 1}$

$\displaystyle \sf{ \implies \: (5 - r) = 2}$

$\displaystyle \sf{ \implies \: r = 3}$

FINAL ANSWER

The required value of r = 3

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