# 7 relatives of a man comprises 4 ladies and 3 gentlemen; his wife has also 7 relatives: 3 of them are ladies and 4โ gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man's relatives and 3 of the wife's relatives?

Navjyot Kalra
9 years ago
Hello Student,
There four possibilities:
(i) 3 ladies from husband’s side and 3 gentlemen from wife’s side.
No. of ways in this case
= 4 C3 x 4 C3 = 4 x 4 = 16
(ii) 3 gentlemen from husband’s side and 3 ladies from wife’s side.
No. of ways in this case = 3 C3 x 3 C 3 = 1 x 1 = 1
(iii) 2 ladies and one gentleman from husband’s side and lady and 2 gentlemen from wife’s side.
No. of ways in this case
= (4 C4 x 3 C1) x (3 C1 x 4 C2) = 6 x 3 x 3 x 6 = 324
(iv) One lady and 2 gentlemen from husband’s side and 2 ladies and one gentlemen from wife’s side.
No. of ways in this case
= (4 C1 x 3 C2) x (3 C2 x 4 C1) = 4 x 3 x 3 x 4 = 1444
Hence the total no. of ways are
= 16 + 1 + 324 + 144 = 485.

Thanks
navjot kalra
SUKRITi c
14 Points
4 years ago
I think it should be 215 as it is 16 +54 +144+1 = 215 and not 16+324+144+1 =ย  485
Your 324ย  which is ( 4C4 x 3C1 ) x ( 4C2 x 3C1 )ย  should it not be (4C2 x 3C1) x (4C2 x 3C1 ) which is 54 ?
ย
Please do correct me if Iโm wrong as I am still an aspirant to make it to IITย  , God Willing.
SUKRITi c
14 Points
4 years ago
I was goingg wrong in calculation. Now I got the corrcet answer 485.
ย
But your (4C4 x 3C1) x (4C2 x 3C1)ย  should be (4C2 x 3C1 ) x (4C2 x 3C1 ) . May be a typographical error in your solutionย  where intended 4C2 was erroneously typed as 4C4 .. With 4C4ย  you will get 54 and not 324
Rishi Sharma
3 years ago
Dear Student,

There four possibilities:
(i) 3 ladies from husband’s side and 3 gentlemen from wife’s side.
No. of ways in this case
= 4 C3 x 4 C3 = 4 x 4 = 16
(ii) 3 gentlemen from husband’s side and 3 ladies from wife’s side.
No. of ways in this case = 3 C3 x 3 C 3 = 1 x 1 = 1
(iii) 2 ladies and one gentleman from husband’s side and lady and 2 gentlemen from wife’s side.
No. of ways in this case
= (4 C4 x 3 C1) x (3 C1 x 4 C2) = 6 x 3 x 3 x 6 = 324
(iv) One lady and 2 gentlemen from husband’s side and 2 ladies and one gentlemen from wife’s side.
No. of ways in this case
= (4 C1 x 3 C2) x (3 C2 x 4 C1) = 4 x 3 x 3 x 4 = 1444
Hence the total no. of ways are
= 16 + 1 + 324 + 144 = 485.

Thanks and Regards