NIMCET 2015 Question Paper with solutions

Question : 25 of 120

Marks: +1, -0

If 42

(\,↖{n}P_2)

↖{n}P_4

then the value of ‘n’ is;

Solution:

We have,

42^nP_2=^nP_4

⇒42{n!}/{(n-2)!}={n!}/{(n-4)!}

⇒{(n-2)!}/{(n-4)!}=42

⇒{(n-2)(n-3)(n-4)!}/{(n-4)!}=42

⇒n^{2}-5n+6=42

n^{2}-5n-36

⇒(n-9)(n+4)=0

n=9

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