Question:
If ${ }^{\prime \prime} \mathrm{C}_{8}={ }^{n} \mathrm{C}_{2}$, find ${ }^{\text {" }} \mathrm{C}_{2}$
Solution:
It is known that, ${ }^{n} C_{a}={ }^{n} C_{b} \Rightarrow a=b$ or $n=a+b$
Therefore,
${ }^{n} \mathrm{C}_{8}={ }^{n} \mathrm{C}_{2} \Rightarrow \mathrm{n}=8+2=10$
$\therefore{ }^{\mathrm{n}} \mathrm{C}_{2}={ }^{10} \mathrm{C}_{2}=\frac{10 !}{2 !(10-2) !}=\frac{10 !}{2 ! 8 !}=\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}=45$
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