26 votes

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.

- $2^{n}$
- $n^{2}$
- $\binom{n}{⌊n/2⌋}^{2}$
- $\binom{2n}{n}$
- None of the above

25 votes

Best answer

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**).

13

**Combinatorial Proof for identity used**

We want to choose $n$ objects from $2n$ objects.

So let's divide this set into two sets of $n$ objects each, call it S1 and S2 respectively.

Now we can choose

$0$ from S1 and $n$ from S2 = $\binom{n}{0} \cdot \binom{n}{n}$

$1$ from S1 and $(n-1)$ from S2 = $\binom{n}{1} \cdot \binom{n}{n-1}$

.

.

.

$n$ from S1 and $0$ from S2 = $\binom{n}{n} \cdot \binom{n}{0}$

adding all up

$\binom{n}{0} \cdot \binom{n}{n} + \binom{n}{1} \cdot \binom{n}{n-1} + \ldots + \binom{n}{n} \cdot \binom{n}{0}$

$= \binom{n}{0} \cdot \binom{n}{0} + \binom{n}{1} \cdot \binom{n}{1} + \ldots + \binom{n}{n} \cdot \binom{n}{n}$

This is a special form of theorem called Vandermonde's identity.

4

^{n}C_{0}.^{n}C_{0} +^{n}C_{1 * }^{n}C_{1}+ ^{n}C_{2}_{ * }^{n}C_{2} + ^{n}C_{3 }_{* }^{n}C_{3}+.......................^{n}C_{n} * ^{n}C_{n } =(^{2n}C_{n})

how u calculate this?

7

**Proof **

**(1+x) ^{n}**=nC0.x

**(x+1) ^{n}**=nC0.x

multiply equation 1 and 2 and compare coefficient of x^{n} in LHS and RHS

**// Coefficient of x ^{r} in (1+x)^{n }is nCr**

so 2nCn=( (nC0)^{2}+(nC1)^{2}+(nC2)^{2}+(nC3)^{2}+......(nCn)^{2 })^{2}

17 votes

Let** ****M:Males**** ****F:Females**

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

CaseSelect no. of M's Out of 3Select no. of F's out of 3Total Ways1 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**

0 votes

0 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

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