Question

Find n, if np6= 20 np3

Answer 1

No real solution is possible for this equation.

1) ⁿPₓ = n!/(n-x)!

2) ⁿP₆ = n!/(n-6)!

3) ⁿP₃ = n!/(n-3)!

4) n!/(n-6)! = 20* n!/(n-3)!

5) n!/(n-6)! * (n-3)!/n! = 20

6) (n-3)!/(n-6)! = 20

7) (n-3)(n-4)(n-5) = 20

 solving this equation gives three values of n

 n^3 - 12 n^2 + 47 n - 80 = 0

n = 6.837, 2.5814 - 2.2443 i,  2.5814 + 2.2443 i

8) Since, we can't have neither decimal or imaginary values in permutation, hence all this values are not possible.

9) If we change 20 in the question to 210, then the value of n which can satisfy this equation would be n= 10.