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 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, is :

  • Option 1)

     468

  • Option 2)

    469

  • Option 3)

    484

  • Option 4)

    485

 

Answers (1)

As we learnt in

Theorem of Combination -

Each of the different groups or selection which can be made by taking r things from n things is called a combination.

^{n}c_{r}=\frac{(n)!}{r!(n-r)!}

- wherein

Where 1\leq r\leq n

 

 We use combinations 

Total number f ways

\Rightarrow\ \; ^{4}C_{0}.^{3}C_{3}.^{3}C_{3}.^{4}C_{0}+\ ^{4}C_{1}.^{3}C_{2}.^{3}C_{2}.^{4}C_{1}+ \ ^{4}C_{2}.^{3}C_{1}.^{3}C_{1}.^{4}C_{2}+ \ ^{4}C_{3}.^{3}C_{0}.^{3}C_{0}.^{4}C_{3}

\Rightarrow\ \; 485

Correct option is 4.

 


Option 1)

 468

This is an incorrect option.

Option 2)

469

This is an incorrect option.

Option 3)

484

This is an incorrect option.

Option 4)

485

This is the correct option.

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