in Combinatory edited by
32 votes

$n$ couples are invited to a party with the condition that every husband should be accompanied by his wife. However, a wife need not be accompanied by her husband. The number of different gatherings possible at the party is

  1. \(^{2n}\mathrm{C}_n\times 2^n\)
  2. \(3^n\)
  3. \(\frac{(2n)!}{2^n}\)
  4. \(^{2n}\mathrm{C}_n\)
in Combinatory edited by


The condition of no one coming is hard to see here.

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5 Answers

73 votes
Best answer

Possible outcome for a couple:

  1. only wife comes
  2. both husband and wife come
  3. neither husband nor wife comes

Thus $3$ possibilities for each couple, so $\underbrace{3\times 3\times 3\times \cdots \times 3}_{\text{n times}}= 3^n$

Correct Answer: $B$

edited by


It would be easy to understand If u give few examples along with the answer...

Thanks in advance

Here it is mentioned wife need not come with husband,

This may give 2 outcomes

wife comes with husband

wife comes with other person other than husband

possible outcomes are

wife comes with husband

wife comes with other person other than husband

husband with wife

and also none

Please clarify @Arjun Sir


@Jarvis You are inventing cases here , which are not there !
wife comes with other person other than husband = Wife not coming with husband !

We are given only 3 cases here !

  1. only wife comes
  2. both come
  3. none come

You might want to include few more cases , if you follow your logic !

Wife comes with Teddy bear :P and so on...

::D Teddy Bear
what is the meaning of none come

if wife is not present with husband  then husband cant come alone in party .. that is the third case in which nobody will come, beacuse their is a strict condition with husband.
consider a couple ,it can be $HW$,$WX$ or $XX$(none) .i.e a couple can be organized in 3 ways and we have $n$ couples .Use Multiplication  theorem $3*3*3*.....*3$ (n times)=$3^{n}$
edited by

The correct answer is (B) 3n

@Jarvis have got some logic !
edited by

Husband are not come alone

Here confusion may arise for the statement : a wife need not be accompanied by her husband. So one can think that wife is allowed to come with somebody else's husband but that would violate the condition every husband should be accompanied by his wife!
@Akash If a husband is in our answer , he should be with His wife only . Clearly mentioned in Question.
0 votes

Might be a long answer but here goes:


let n = 2 (h1 - w1, h2 - w2)


Different Possibilities:


A) No one

- {}

B) One wife

- w1

- w2

C) Both wives

- w1, w2 

D) One husband (therefore corresponding wife lol)

- h1,w1

- h2,w2

E) Both husbands (therefore both wives)

- h1,h2,w1,w2

F) One husband (with corresponding wife) + 1 wife

- h1,w1,w2

- h2,w2,w1


There are 9 different possibilities, only option C matches.


0 votes

The possibilities to attend party is
i) Both husband and wife comes
ii) Only wife comes
iii) Both are not come
The no. of different gatherings possible at party is
= 3 * 3 * 3 * 3 * ... n times
= 3n

Option B

0 votes
We can also think in terms of boolean inputs.

Husband            Wife              Condition

0                         0                   both don’t come                                                 πŸ‘

0                         1                   wife without husband                                        πŸ‘

1                         0                   husband cannot come by himself                   πŸ‘Ž

1                         1                   both come together                                           πŸ‘

Thus 3 valid cases!

Answer is 3^n.
0 votes
Given n couples means n husband and n wife.

Let i = number of husband coming to party

Then along with him wife should also accompany.

So remaining number of wife = n – i

So, remaining n-i wife has 2 choices, either she comes to party or don’t.

${\binom{n}{i}}$ = Choosing i husband out of n that come to party

$2^{n-i}$  = 2 choices for remaining n-i wife

So equation reduces to : ( Take summation over all i's)

 = $\sum_{1}^{n}$  ${\binom{n}{i}}$ * $2^{n-i}$

 = $\sum_{1}^{n}$  ${\binom{n}{i}}$ * ($\frac{2^{n}}{2^{i}}$ )

 = $2^{n}$ $\sum_{1}^{n}$  ${\binom{n}{i}}$ / ($2^{i}$ )

 = $2^{n}$ ( $\frac{{\binom{n}{0}}} {2^{0}} $ + $\frac{{\binom{n}{1}}} {2^{1}}$  + $\frac{{\binom{n}{2}}} {2^{2}}$  β€¦β€¦..   +  $\frac{{\binom{n}{n}}} {2^{n}})$)

 = $2^{n}$ $\frac{ (  {\binom{n}{0}} * (2^{n})  +  {\binom{n}{1}} * (2^{n-1})  +  {\binom{n}{2}} * (2^{n-2})  +  .………  +  {\binom{n}{n}} * (2^{0}) )}{2^{n}} $

Using Binomial Identity:

$(a+b)^{n}$ = $\sum_{1}^{n} \binom{n}{i} a^{i} b^{n-i}$

Put a =1 and b = 2, we get

$3^{n}$  = ( $ {\binom{n}{0}} * (2^{n})  +  {\binom{n}{1}} * (2^{n-1})  +  {\binom{n}{2}} * (2^{n-2})  +  .………  +  {\binom{n}{n}} * (2^{0}) )$


Finally putting value of above identity into equation,

 = $2^{n}$ $\frac{ (  {\binom{n}{0}} * (2^{n})  +  {\binom{n}{1}} * (2^{n-1})  +  {\binom{n}{2}} * (2^{n-2})  +  .………  +  {\binom{n}{n}} * (2^{0}) )}{2^{n}} $

= $2^{n}$ ($ \frac{3^{n}}{2^{n}})  $

= $3^{n}$

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