(3) If 2np3 = 100× np2
Permutation
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2
Answer:
A]
2n
P
3
=100×
n
P
2
=
(2n−3)!
(2n)!
=100×
(n−2)!
n!
=
(2n−3)!
(2n−3)!(2n−2)(2n−1)2n
=
(n−2)!
100×(n−2)!(n−1)n
=2(2n-2)(2n-1)=100(n-1)
=2(4n
2
−6n+2)=100n−100
=8n
2
−12n+4=100n−100
=8n
2
−112n+104=0
=4n
2
−56n+52=0
=2n
2
−28n+26=0
==n
2
−14n+13=0
= n (n-13) - 1 (n-13) = 0
n = 1, n = 13
∴ n = 13 [as n=1 not possible
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