A) [8, 9]
B) [10, 12)
C) (11, 13]
D) (14, 17)
Correct Answer: B
Solution :
Let no. of men = n No. of women = 2 Total participants = n + 2 No. of games that\[{{M}_{1}}\]plays with all other men = 2(n ? 1) These games are played by all men \[{{M}_{2}},{{M}_{3}},\] ......, \[{{M}_{n}}.\] So, total no. of games among men = n.2(n ? 1). However, we must divide it by ?2?, since each game is counted twice (for both players). So, total no. of games among all men = n(n ? 1) ....... (1) Now, no. of games \[{{M}_{1}}\]plays with \[{{W}_{1}}\]and \[{{W}_{2}}=4\](2 games with each) Total no. of games that \[{{M}_{1}},{{M}_{2}},\] ....., \[{{M}_{n}}\] play with \[{{W}_{1}}\]and \[{{W}_{2}}=4n\] .... (2) Given : \[n(n-1)-4n=66\]\[\Rightarrow \]\[n=11-6\] As the number of men can't be negative. So, n = 11
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