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 ?
this is similar to handshake problem , if there are nn people , then number of handshakes = 1 + 2 + . . . n = n ∗ ( n − 1 ) / 2 no . of women be xx and given x ∗ ( x − 1 ) / 2 = 45 this gives x = 10 | | | y no . of men : solve y ∗ ( y − 1 ) / 2 = 190 . this gives y = 20 so total men vs . women = 10 * 20 = 200 . answer is d
multiply(divide(add(const_1, sqrt(add(const_1, multiply(multiply(45, const_2), const_4)))), const_2), divide(add(const_1, sqrt(add(const_1, multiply(multiply(190, const_2), const_4)))), const_2))