Question

if np3=120 then n will be
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Answer 1

Answer:

n=6

Step-by-step explanation:

np3=120

n!/(n-3)! =120

[n(n-1)(n-2)(n-3)!]/(n-3)! =120

n(n-1)(n-2)=120

n^3 - 3n^2 + 2n = 120

n^3 - 3n^2 + 2n - 120 = 0

n^3 - 6n^2 + 3n^2 + 2n - 120 = 0

n^2(n-6) + 3n(n-6) + 20(n-6) = 0

(n-6)×(n^2 + 3n + 20) = 0

n-6=0 .... eq(i)

n=6

and

n^2 + 3n + 20 = 0. ......eq(ii)

according to eq 2nd n does not belong to real number

hence n=6 answer