Combination - KEY QUESTION

**cnmath16** · November 23rd 2008, 02:00 PM

Solve for n if P(n,3) = 2C (n,2)

**Shyam** · November 23rd 2008, 03:43 PM

Originally Posted by cnmath16

Solve for n if P(n,3) = 2C (n,2)

P(n, 3) = 2 C(n, 2)

$\frac{n!}{(n-3)!}=2\frac{n!}{(n-2)! \;2!}$

$n(n-1)(n-2)=2\frac{n(n-1)}{2\times 1}$

$n(n-1)(n-2)-n(n-1)=0$

$n(n-1)[(n-2)-1]=0$

$n(n-1)(n-3)=0$

n=0, 1, 3

since, in P(n, r), $n\ge r$

therefore, $n\ge 3$

so, n = 3

did you get it now???

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