Explanation

\(^{5y}{C}_2\) = 190

 \(\frac{5y(5y-1) (5y-2)!}{(5y-2)!(2)!}\) = 190

 \(\frac{5y(5y-1)}{2}\) = 190

25y\(^2\) - 5y = 380

5y\(^2\) - 5y - 76 = 0

 = (y-4) (5y+ 19) = 0

Either y - 4 = 0 or 5y+ 19 = 0;

y = 4 or -19/5

: y= 4

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