Question

a man has 7 relatives ,4 of them are ladies and 3 gentleman ,his wife also has 7 relatives ,3 of them are ladies and 4 gentleman .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 wifes relative is?

Answer 1

@ganeshie8

Answer 2

@ParthKohli

Answer 3

now you count each of the 4 cases individually

Answer 4

7C3 * 7C3 is total ways to get 3 from man and womans side
out of these total ways how many with 3 women and 3 men

Answer 5

7C0women
7C1womn
7C2Woman

u wanna remove these cases

Answer 6

(7C3)^2-7C0-7C1-7C2

Answer 7

1196?

Answer 8

lets call the original man A , and his wife B
There are 4 cases

Let w = woman, m = male

if A gets 3 w relatives
B must get 3 m relatives

this is 4C3 * 4C3


if A gets 2 w , 1m
B gets 2m, 1 w
this is (4C2*3C1)*(4C2 * 3C1)



if A gets 1 w , 2 m
B gets 1m, 2w

this is (4C1*3C2)*(4C1*3C2)


if A gets 3 m
B gets 3 w

this is (3C3) *(3C3)

Answer 9

answer is 485

Answer 10

oh i know okay

Answer 11

yes my method works

Answer 12

4C3 * 4C3 + (4C2*3C1)*(4C2 * 3C1) + (4C1*3C2)*(4C1*3C2) + (3C3) *(3C3) = 485

Answer 13

perl is right i think

Answer 14

do you see how i got those numbers?

Answer 15

how?

Answer 16

let A = original husband, B = original wife

Answer 17

\(\sum_{i=0}^3 (^4_i)(^3_{3-i})(^3_i)(^4_{3-i})=485\)

Answer 18

you need there to be 3 relatives from A, and relatives from B , under the constraint that there are 3 women and 3 men. and so i went through the 4 cases.
for example the first case
if A gets 3 w relatives
B must get 3 m relatives

A has 4 women to choose from, and you pick 3 women.
B has 4 men to choose from, and you pick 3 men.
So this gives us , by multiplication rule
4C3 * 4C3

Answer 19

dan;s way is cool as well. he first overcounted then subtracted

Answer 20

thanks a lot.

Answer 21

hey why did you delete your work @dan815 I was studying it :)

Answer 22

mathmates solution makes a nice sum expression for this