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Two American, two British, Two Canadian, Two Dutch and two Egyptians are present in a dance club. Number of ways to make 5 pairs such that people of same nationality are not paired is N then N100 is

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answer is 5.44.

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Detailed Solution

A = American are pairedB = British are paired Total =10!25⋅15!n( total )−∑n(A)+∑n(A∩B)………….9.7.5.3−5.(_7.5.3)+10(5.3)−10(3)+5(1)−1_=544

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