Question:
If ${ }^{(n 2-n)} C_{2}={ }^{(n 2-n)} C_{4}=120$ then find the value of $n .$
Solution:
Given: ${ }^{(n 2-n)} \mathrm{C}_{2}={ }^{(n 2-n)} \mathrm{C}_{4}=120$ Need to find: Value of $n{ }^{(n 2-n)} \mathrm{C}_{2}={ }^{(n 2-n)} \mathrm{C}_{4}=$
120 We know, one of the property of combination is: If ${ }^{n} C_{r}={ }^{n} C_{t}$, then, (i) $r=t$ OR (ii) $r+$ $t=n$ Applying property (ii) we get, $n^{2}-n=2+4=6 n^{2}-n-6=0 n^{2}-3 n+2 n-6=$ $0 n(n-3)+2(n-3)=0(n-3)(n+2)=0$ So, the value of $n$ is either 3 or $-2$