Thread: Immediate help needed!Exams tomorrow!

I am unable to solve this question at all.

In how many ways can a lawn tennis mixed double be made up from seven married couples if no husband and wife can play in the same set?

Answer is 840.

This question is from the chapter Permutations(using only fundamental principle of multiplication, addition and formula of permutations, NO combinations)

Can anyone tell me how is this solved?

(I marked the husbands as h1,h2,h3,h4,h5,h6,h7. h1 can get 1 from 6 partners, h2 can get 1 from 5 and so one. But this approach failed)

7 choices of first male, 6 choices for his partner (cannot choose his wife)

5 choices for second male ( cannot choose husband of first female out of remaining six men ) , 4 choices for second female ( of six remaining women exclude wives of two selected males )

so 7*6*5*4=840