Question

A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then what is the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party?
${\text{A}}{\text{. 468}} \\\ {\text{B}}{\text{. 469}} \\\ {\text{C}}{\text{. 484}} \\\ {\text{D}}{\text{. 485}} \\\$

Answer 1

Hint: Here, we will proceed by finding the number of ways of inviting all the 3 ladies are from X’s friend list and all the 3 men are from Y’s friend list, inviting 2 ladies and 1 man are from X’s friend list and 1 lady and 2 men are from Y’s friend list, inviting 1 lady and 2 men are from X’s friend list and 2 ladies and 1 man are from Y’s friend list and inviting all the 3 men are from X’s friend list and all the 3 ladies are from Y’s friend list separately.

Complete step-by-step answer:

Given, man X has 7 friends out of which 4 are ladies and 3 are men and his wife Y has 7 friends out of which 3 are ladies and 4 are men where X and Y have no common friend.
As we know that the number of ways of selecting r items out of n items is given by ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ where $0! = 1$.
Following cases can be made for inviting 3 ladies and 3 men to a party so that 3 friends of each of X and Y are there.
Case-I: When all the 3 ladies are from X’s friend list and all the 3 men are from Y’s friend list.
Number of ways = Number of ways of selecting 3 ladies out of 4 ladies$\times$Number of ways of selecting 3 men out of 4 men
$\Rightarrow$Number of ways $= {}^4{C_3} \times {}^4{C_3} \\\ = \left( {\dfrac{{4!}}{{3!\left( {4 - 3} \right)!}}} \right) \times \left( {\dfrac{{4!}}{{3!\left( {4 - 3} \right)!}}} \right) \\\ = \left( {\dfrac{{4.3!}}{{3!1!}}} \right) \times \left( {\dfrac{{4.3!}}{{3!1!}}} \right) \\\ = 4 \times 4 \\\ = 16 \\\$.
Case-II: When 2 ladies and 1 man are from X’s friend list and 1 lady and 2 men are from Y’s friend list.
Number of ways = Number of ways of selecting 2 ladies out of 4 ladies$\times$Number of ways of selecting 1 man out of 3 men $\times$Number of ways of selecting 1 lady out of 3 ladies$\times$Number of ways of selecting 2 men out of 4 men
$\Rightarrow$Number of ways

= {}^4{C_2} \times {}^3{C_1} \times {}^3{C_1} \times {}^4{C_2} \\\ = {\left[ {{}^4{C_2} \times {}^3{C_1}} \right]^2} \\\ = {\left[ {\left( {\dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}} \right) \times \left( {\dfrac{{3!}}{{1!\left( {3 - 1} \right)!}}} \right)} \right]^2} \\\ = {\left[ {\left( {\dfrac{{4.3.2!}}{{2.1!2!}}} \right) \times \left( {\dfrac{{3.2!}}{{1!2!}}} \right)} \right]^2} \\\ = {\left[ {\left( 6 \right) \times \left( 3 \right)} \right]^2} \\\ = {\left[ {18} \right]^2} \\\ = 324 \\\ $$. Case-III: When 1 lady and 2 men are from X’s friend list and 2 ladies and 1 man are from Y’s friend list. Number of ways = Number of ways of selecting 1 lady out of 4 ladies$ \times $Number of ways of selecting 2 men out of 3 men $ \times $Number of ways of selecting 2 ladies out of 3 ladies$ \times $Number of ways of selecting 1 man out of 4 men $ \Rightarrow $Number of ways

= {}^4{C_1} \times {}^3{C_2} \times {}^3{C_2} \times {}^4{C_1} \\
= {\left[ {{}^4{C_1} \times {}^3{C_2}} \right]^2} \\
= {\left[ {\left( {\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}}} \right) \times \left( {\dfrac{{3!}}{{2!\left( {3 - 2} \right)!}}} \right)} \right]^2} \\
= {\left[ {\left( {\dfrac{{4.3!}}{{3!}}} \right) \times \left( {\dfrac{{3.2!}}{{2!1!}}} \right)} \right]^2} \\
= {\left[ {\left( 4 \right) \times \left( 3 \right)} \right]^2} \\
= {\left[ {12} \right]^2} \\
= 144 \\

Case-IV: When all the 3 men are from X’s friend list and all the 3 ladies are from Y’s friend list. Number of ways = Number of ways of selecting 3 men out of 3 men$ \times $Number of ways of selecting 3 ladies out of 3 ladies $ \Rightarrow $Number of ways $ = {}^3{C_3} \times {}^3{C_3} \\\ = {\left[ {{}^3{C_3}} \right]^2} \\\ = {\left[ {\left( {\dfrac{{3!}}{{3!\left( {3 - 3} \right)!}}} \right)} \right]^2} \\\ = {\left[ {\left( {\dfrac{1}{{0!}}} \right)} \right]^2} \\\ = {\left[ {\left( {\dfrac{1}{1}} \right)} \right]^2} \\\ = 1 \\\ $. For the required number of ways, we will add all the number of ways obtained in the above four cases. So, the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party = 16+324+144+1=485. Therefore, the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party are 485 ways. Hence, option D is correct. Note: In this particular problem, we will add the number of ways obtained in all the above four different cases in order to get the required total number of ways because any of these four cases can occur. Also, here we have to select the persons who will be getting the invitations and for selection we use combinations whereas for arrangements we use permutations.