Math, asked by Anonymous, 6 months ago

Find n if n^p^3 = 60.
n^p^3=60 (given
(n-3)
пP3​

Answers

Answered by tanavnaidu06
0

Answer:

Given,

P

5

n

= 20 P

3

n

(n−5)!

n!

= 20

(n−3)!

n!

(n−5)!

n!

= 20

(n−3)(n−4)(n−5)!

n!

1 = 20

(n−3)(n−4)

1

(n−3)(n−4)=20

n

2

−7n+12=20

n

2

−7n+12−20=0

n

2

−7n−8=0

n

2

+n−8n−8=0

n(n+1)−8(n+1)=0

(n−8)(n+1)=0

n=8,−1

Since , n can not be nagative . n

= −1

Hence, n=8.

Answered by diptimayeejena0706
1

Answer:

n=5

Step-by-step explanation:

Given, n^p^3 = 60

So, n^p^3 = 60

or, n!/(n-3)! = 60

or, n(n-1)(n-2)(n-3)!/(n-3)! = 60

or, n(n-1)(n-2) = 60

or, n(n-1)(n-2) = 5×4×3

or, n = 5

As n= 5

So, (n-3) = (5 - 3) = 2

And, n^p^3 = n!/(n-3)!

= 5!/(5-3)! [Putting the value of n = 5]

= 2!×3×4×5/2!

= 3×4×5

= 60

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