permutations and combinations

**Twinkie** · January 1st 2013, 08:18 AM

Ok I have been having some trouble with thease questions. Could you show me step by step how to do them?

1. Solve for n.
n-1 C 2 = 6

2. Solve for the variable.
n P 3 =720

**Soroban** · January 1st 2013, 08:39 AM

Hello, Twinkie!

Do you understand the meaning of those symbols?

$\text{1. Solve for }n\!:\;_{n-1}C_2 \,=\, 6$

We have: . $\frac{(n-1)!}{2!(n-3)!} \:=\:6 \quad\Rightarrow\quad \frac{(n-1)(n-2)}{2} \:=\:6$

. . $(n-1)(n-2) \:=\:12 \quad\Rightarrow\quad n^2 - 3n + 2 \:=\:12$

. . $n^2 - 3n - 10 \:=\:0 \quad\Rightarrow\quad (n-5)(n+2) \:=\:0$

Therefore: . $n = 5,\;{\color{red}\rlap{//////}}n=\text{-}2$

$\text{2. Solve for }n\!:\;_n P_3 \,=\,720$

We have: . $\frac{n!}{(n-3)!} \:=\:720 \quad\Rightarrow\quad n(n-1)(n-2) \:=\:720$

. . $n^3 - 3n^2 + 2n - 720 \:=\:0 \quad\Rightarrow\quad (n-10)(n^2+7n+72) \:=\:0$

Therefore: . $n \,=\,10$

**Twinkie** · January 1st 2013, 09:23 AM

Ok, thankyou for the help!

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