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08 May 2026, 22:11
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
27% (02:16) correct
73%
(02:27)
wrong
based on 15
sessions
History
Date
Time
Result
Not Attempted Yet
An English-language class has 8 students: one man and one woman from each of four countries. The students are to be divided into 4 pairs so that no pair contains two students of the same gender or from the same country. How many distinct pairings are possible?
A. 4
B. 6
C. 7
D. 8
E. 9
A. 4
B. 6
C. 7
D. 8
E. 9
08 May 2026, 23:11
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I tried approaching this by thinking in terms of pairing men and women from different countries, but I’m still not seeing a clean or fast way to count the total distinct pairings without getting messy.
Could you please explain the shortest or most efficient way to think about this problem? I feel like there’s probably a neat pattern or setup here that I’m missing.
Would really appreciate a structured breakdown of the logic.
— Rajdeep
Could you please explain the shortest or most efficient way to think about this problem? I feel like there’s probably a neat pattern or setup here that I’m missing.
Would really appreciate a structured breakdown of the logic.
— Rajdeep
10 May 2026, 10:09
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kevincan
\(c_1: m_1, w_1\)
\(c_2: m_2,w_2\)
\(c_3: m_3,w_3\)
\(c_4: m_4,w_4\)
\(m_1*3 = 3\) ( After choosing \(m_1\) we have \(3\) options for woman)
\(m_2*3 = 3\) ( After choosing \(m_2\) we have \(3\) options for woman)
\(m_3*1 = 1\) ( After choosing \(m_3\) we have only \(1\) option for woman)
\(m_4*1 = 1\) ( After choosing \(m_4\) we have only \(1\) option for woman)
Total ways : \(3*3*1*1= 9 \)
Ans E
Hope it helped.
