Question

7 relatives of a man comprises 4 ladies and 3 gentleman his wife has also 7 relatives, 3 of them are ladies and 4 gentleman. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of the man's relatives and 3 of the wife’s relatives?
${\text{a}}{\text{ 485}} \\\ {\text{b}}{\text{ 500}} \\\ {\text{c}}{\text{ 486}} \\\ {\text{d}}{\text{ 102}} \\\$

Answer 1

Hint: Use the property of combination which is ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}$. There are four possible cases. We can find out the total number of ways by adding results of all four cases.

Complete step-by-step answer:
A man has 7 relatives (4 ladies and 3 gentlemen).
His wife also has 7 relatives (3 ladies and 4 gentlemen).
Now they have to invite (3 ladies and 3 gentlemen, so that there are 3 of the man's relatives and 3 of the wife’s relatives.
The possible cases are:
Case 1: A man invites 3 ladies and his wife invites 3 gentlemen.
For man out of his 7 relatives in which 4 are ladies and he has to invite 3 ladies so we use combination rule $= {}^4{C_3}$
For his wife out of his 7 relatives in which 4 are gentlemen and she has to invite 3 gentlemen so we use combination rule $= {}^4{C_3}$
So we have to multiply these two above conditions to get the number of ways in which a man invites 3 ladies and his wife invites 3 gentlemen.
$\Rightarrow {}^4{C_3} \times {}^4{C_3}$
Now as we know ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}$
$\Rightarrow {}^4{C_3} \times {}^4{C_3} = \left( {\dfrac{{4!}}{{\left( {4 - 3} \right)! \times 3!}}} \right)\left( {\dfrac{{4!}}{{\left( {4 - 3} \right)! \times 3!}}} \right) = 4 \times 4 = 16$
Case 2: A man invites (2 ladies, 1 gentlemen) and his wife invites (2 gentlemen, 1 lady)
For man out of his 7 relatives in which 4 are ladies and 3 gentlemen and he has to invite 2 ladies and 1 gentlemen so we use combination rule $\left( {{}^4{C_2} \times {}^3{C_1}} \right)$
For his wife out of his 7 relatives in which 4 are gentlemen and 3 ladies and she has to invite 2 gentlemen and 1 lady so we use combination rule $\left( {{}^3{C_1} \times {}^4{C_2}} \right)$
So we have to multiply these two above conditions to get the number of ways in which a man invites 2 ladies and 1 gentleman and his wife invites 2 gentlemen and 1 lady.
$\Rightarrow \left( {{}^4{C_2} \times {}^3{C_1}} \right)\left( {{}^3{C_1} \times {}^4{C_2}} \right) = \left( {\left( {\dfrac{{4!}}{{2! \times 2!}}} \right)\left( {\dfrac{{3!}}{{2! \times 1!}}} \right)} \right)\left( {\left( {\dfrac{{3!}}{{2! \times 1!}}} \right)\left( {\dfrac{{4!}}{{2! \times 2!}}} \right)} \right) = 324$
Case 3: A man invites (1lady, 2 gentlemen) and his wife invites (1 gentlemen, 2 ladies)
For man out of his 7 relatives in which 4 are ladies and 3 gentlemen and he has to invite 1 lady and 2 gentlemen so we use combination rule $\left( {{}^4{C_1} \times {}^3{C_2}} \right)$
For his wife out of his 7 relatives in which 4 are gentlemen and 3 ladies and she has to invite 1 gentlemen and 2 ladies so we use combination rule $\left( {{}^3{C_2} \times {}^4{C_1}} \right)$
So we have to multiply these two above conditions to get the number of ways in which a man invites 1 lady and 2 gentlemen and his wife invites 1 gentleman and 2 ladies.
$\Rightarrow \left( {{}^4{C_1} \times {}^3{C_2}} \right)\left( {{}^3{C_2} \times {}^4{C_1}} \right) = \left( {\left( {\dfrac{{4!}}{{3! \times 1!}}} \right)\left( {\dfrac{{3!}}{{1! \times 2!}}} \right)} \right)\left( {\left( {\dfrac{{3!}}{{1! \times 2!}}} \right)\left( {\dfrac{{4!}}{{3! \times 1!}}} \right)} \right) = 144$
Case 4: A man invites 3 gentlemen and his wife invites 3 ladies
For man out of his 7 relatives in which 3 are gentlemen and he has to invite 3 gentlemen so we use combination rule $= {}^3{C_3}$
For his wife out of his 7 relatives in which 3 are ladies and she has to invite 3 ladies so we use combination rule $= {}^3{C_3}$
So we have to multiply these two above conditions to get the number of ways in which a man invites 3 gentlemen and his wife invites 3 ladies.
$\Rightarrow {}^3{C_3} \times {}^3{C_3} = \left( {\dfrac{{3!}}{{0! \times 3!}}} \right)\left( {\dfrac{{3!}}{{0! \times 3!}}} \right) = 1$
Therefore total number of ways are the addition of all the above cases
${\text{ = }}16 + 324 + 144 + 1 = 485$
Hence, option (a) is correct.

Note: In such types of questions students may forget that man invites guests then man himself should be excluded. Due to this mistake, students get the wrong solution.