Question

If 6![1/2! +4/3!]=7pr find r

Answer 1

Answer:

1 is the value of r .....

Answer 2

The value of r would be 4

Step-by-step explanation:

Given,

$6!(\frac{1}{2!}+\frac{4}{3!})=^7p_r$

We know that,

$^nP_r=\frac{n!}{(n-r)!}$

Also, n! = n(n-1)(n-2).......3.2.1

Where, n = natural number.

So,

$6!(\frac{1}{2}+\frac{4}{3\times 2\times 1})=\frac{7!}{(7-r)!}$

$6!(\frac{1}{2}+\frac{2}{3})=\frac{7!}{(7-r)!}$

$6!(\frac{3+4}{6})=\frac{7!}{(7-r)!}$

$6!(\frac{7}{6})=\frac{7!}{(7-r)!}$

$\frac{7!}{6}=\frac{7!}{(7-r)!}$

By comparing,

(7-r)! = 6

∵ 3! = 6

⇒ (7-r)! = 3!

⇒ 7-r = 3

⇒ r = 7 - 3 = 4

Hence, the value of r is 7.

#Learn more:

7Pr =42, find the value of r

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