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$n$ 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. \(^{2n}\mathrm{C}_n\)
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5 Answers

Best answer
89 votes
89 votes

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
3 votes
3 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}} $ + $\frac{{\binom{n}{1}}} {2^{1}}$  + $\frac{{\binom{n}{2}}} {2^{2}}$  ……..   +  $\frac{{\binom{n}{n}}} {2^{n}})$)

 = $2^{n}$ $\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}})  $

= $3^{n}$
1 votes
1 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
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.

 

Answer:

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