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Probability Level 3

In a chess competition, y y men and x x women were present. Each player played with other exactly once. The matches between man and man were 10 and between woman and woman were 21. If matches between man and woman were z z , find the value of x + y + z x+y+z .

None of these 47 55 44 53 39

Prince Loomba by Prince Loomba , 21, India

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2 solutions

Chew-Seong Cheong
Oct 10, 2016
  • The number of matches between men = ( y 2 ) = y ( y − 1 ) 2 = 10 = \displaystyle {y \choose 2} = \dfrac {y(y-1)}2 = 10 ⟹ y ( y − 1 ) = 20 \implies y(y-1) = 20 ⟹ y = 5 \implies y = 5
  • The number of matches between women = ( x 2 ) = x ( x − 1 ) 2 = 21 = \displaystyle {x \choose 2} = \dfrac {x(x-1)}2 = 21 ⟹ x ( x − 1 ) = 42 \implies x(x-1) = 42 ⟹ x = 7 \implies x = 7
  • Total number of matches N = ( 12 2 ) = 12 × 11 2 = 66 N = \displaystyle {12 \choose 2} = \dfrac {12\times 11}2 = 66
  • The number of matches between man and women z = N − 10 − 21 = 66 − 10 − 21 = 35 z = N - 10-21 = 66-10-21 = 35

⟹ x + y + z = 7 + 5 + 35 = 47 \implies x+y+z = 7+5+35 = \boxed{47}

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Prince Loomba
Oct 10, 2016

xC2=21 implies x=7, yC2=10 implies y=5. Man-woman matches=5C1×7C1=35. So answer 5+7+35=47

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