Find r if 5Pr = 6Pr-1

Asked by Abhisek | 1 year ago |  72

##### Solution :-

$$\dfrac{5!}{(5-r)!}= \dfrac{6!}{(6-r+1)}$$

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

$$\dfrac{5!}{(5-r)!}= \dfrac{6}{(7-r)(6-r)(5-r)!}$$

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

= (7 - r)(6- r)=12

= 42-6r -7r +r2=12

= (7 - r)(6- r)=6

42-6r -7r +r =6

r2 -13r +36= 0

r2 -4r -9r(r - 4)=0

(r - 4)(r -9)=0

= r-4=0 or r = -9 = 0

= r =4 or r =9

It is known that

$$^6P_{r-1}=\dfrac{n!}{(n-r)!}$$'

where $$0\leq r\leq n$$

$$0\leq r\leq 5$$

r = 4

Answered by Abhisek | 1 year ago

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