Question:
If ${ }^{m} C_{1}={ }^{n} C_{2}$ prove that $m=\frac{1}{2} n(n-1)$
Solution:
Given: ${ }^{m} C_{1}={ }^{n} C_{2}$ Need to prove: $m=\frac{1}{2} n(n-1){ }^{m} C_{1}={ }^{n} C_{2} \Rightarrow \frac{m !}{1 !(m-1) !}=\frac{n !}{2 !(n-2) !} \Rightarrow$
$\frac{m(m-1) !}{(m-1) !}=\frac{1}{2} \frac{n(n-1)(n-2) !}{(n-2) !} \Rightarrow m=\frac{1}{2} n(n-1)$ [Proved]
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