Question
If ${ }^{n} C_{2}=45$, find the value of $n$.
✓
Answer
$ { }^{n} \mathrm{C}_{r}=\frac{n !}{r !(n-r) !}$
$\therefore { }^{n} \mathrm{C}_{2}=\frac{n !}{2 !(n-2) !}$
$\therefore 475=\frac{n(n-1)(n-2) !}{2 \times 1 \times(n-2) !}$
$\therefore 90=n(n-1)$
$\therefore n(n-1)=10(10-1)$
$\therefore n=10 $
$\therefore { }^{n} \mathrm{C}_{2}=\frac{n !}{2 !(n-2) !}$
$\therefore 475=\frac{n(n-1)(n-2) !}{2 \times 1 \times(n-2) !}$
$\therefore 90=n(n-1)$
$\therefore n(n-1)=10(10-1)$
$\therefore n=10 $
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