Question

A man has seven relative 4 of them are ladies and 3 gentlemen, his wife also has 7 relatives out of which 3 are ladies and 4 gentlemen. The number of ways in which they can invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of the man's relative and 3 of the wife's relative are

• A. 144
• B. 340
• C. 484
• D. 485

Answer 1

Answer: (D)

485

There are 4 Mutually exclusive cases in which three from the man's relative and 3 from wife's relative can be invited.
These cases are given below by column A,B,C,D respectively corresponds to man's men relaive man's women relative, wife's men relative,wife's women relative.
No.fo realtive $\begin{array}{cccc}\left(3\right)& \left(4\right)& \left(4\right)& \left(3\right)\\ A& B& C& D\\ 0& 3& 3& 0\\ 1& 2& 2& 1\\ 2& 1& 1& 2\\ 3& 0& 0& 3\end{array}$
Number of ways ${\Rightarrow }^3{C}_0{\times }^4{C}_3{\cdot }^4{C}_3{\cdot }^3{C}_0=16$ ${\Rightarrow }^3{C}_1{\cdot }^4{C}_2{\cdot }^4{C}_2{\cdot }^3{C}_1=324$
${\Rightarrow }^3{C}_2{\cdot }^4{C}_1{\cdot }^4{C}_1{\cdot }^3{C}_2=144$ ${\Rightarrow }^3{C}_3{\cdot }^4{C}_0{\cdot }^4{C}_0{\cdot }^3{C}_3=1$
$\therefore$ Total required number are $S=485$