In a chess competition involving some men and women, every player needs to play exactly one game with every other player. It was found that in 45 games, both the players were women and in 190 games, both players were men. What is the number of games in which one person was a man and other person was a woman?
Let total number of women =w
total number of men =m
Number of games in which both players were women =45
=>$wc_2 $= 45
=>$\dfrac{w(w-1)}{2}$=45
=>w(w-1)=90
=>w=10
Number of games in which both players were men =190
=>mc_2 = 190
=>$\dfrac{m(m-1)}{2}$=190
=>m(m-1)=380
=>m=20
We have got that
Total number of women = 10
Total number of men = 20
Required number of games in which one person was a man and other person was a woman
$^{20}C_1 × ^{10}C_1$ =20×10=200