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There is a set of $2n$ people: $n$ male and $n$ female. A good party is one with equal number of males and females (including the one where none are invited). The total number of good parties is.

  1. $2^{n}$
  2. $n^{2}$
  3. $\binom{n}{⌊n/2⌋}^{2}$
  4. $\binom{2n}{n}$
  5. None of the above
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4 Answers

Best answer
27 votes
27 votes

There are $n$ men and $n$ women

Now we can select $1$ woman from $n$ in $^{n}C_{1}$

Same way $1$ man can be selected  $^{n}C_{1}$ ways

So, for $1$ woman and $1$ man we can get  $^{n}C_{1}\times ^{n}C_{1}$ ways $\qquad \to (1)$

Similarly, we can select $2$ woman from $n$ women in $^{n}C_{2}$

$2$ man can be selected in  $^{n}C_{2}$ ways

So, for $2$ woman and $2$ man we can get  $^{n}C_{2}\times ^{n}C_{2}$ ways $\qquad \to (2)$

$\vdots$

For $n$ woman and $n$ man we can get   $^{n}C_{n}\times ^{n}C_{n}$ ways$\qquad \to (n)$

Now, by adding these equations $(1),(2), \dots, (n)$ we get ,

$^{n}C_{0}\times ^{n}C_{0} + ^{n}C_{1}\times ^{n}C_{1}+ ^{n}C_{2}\times ^{n}C_{2} + ^{n}C_{3}\times ^{n}C_{3}+\ldots +^{n}C_{n}\times ^{n}C_{n} =\left(^{2n}C_{n}\right)$

Hence, Ans will be (D).

edited by
19 votes
19 votes

Let M:Males F:Females

Suppose n=3, then there are total 2n persons; M=F=3

Case Select no. of M's Out of 3 Select no. of F's out of 3 Total Ways
1 0 0 3C0*3C0=1
2 1 (select 1 M out of 3 so 3C1 ways) 1 3C1*3C1=9
3 2 2 3C2*3C2=9
4 3 3 3C3*3C3=1
    Total 20

Which fits none of the above options but this is equals to 2nCn=6C3=20

1 votes
1 votes
Alternate approach, solve using example.

For n=2 and there are 2n people = 4
Consider that the four people are {M1, M2, F1, F2} – M stands for male and F stands for Female.

Now, let's count the number of good parties. Considering that k is the number of males or females.

for k=0, number of good parties = {} = 1
for k=1, number of good parties = {M1, F1}, {M1, F2}, {M2, F1}, {M2, F2} = 4
for k=2, number of good parties = {M1, M2, F1, F2} = 1

total number of good parties = 1 + 4 + 1 = 6

Comparing against the answers derived from options
option 1 = 2^n = 4 is incorrect
option 2 = n^2 = 4 is incorrect
option 3 = (n choose n/2)^2 = 4 is incorrect
option 4 = (2n choose n) = 6 is correct

Ans: D
0 votes
0 votes
we have to select n females from 2n people

=$\binom{2n}{n}$

so ans is D
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