# TIFR2015-A-8

1.9k views

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.

edited
0

Please Note The Correct Question :

( @Shaik Masthan please edit the options ) 0
edited !

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
1

How is 2nCn equal to  2n/n

3
you did not included the case when there is no party...
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  =  $\binom{n}{n} \cdot \binom{n}{0}$

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

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

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

0
@srestha it should be 2ncn- nc0.nc0 rt?
0

Why nC0.nC0 should not be there? It also be included in calculation

1
but you did not mention in your ans sequence ?
1
yes, corrected. chk it by @Rajesh example :)
0
yeah :)
2
the option D is not 2n/n. It is C(2n,n).
2

nC0.nC0 +nC1 * nC1+  nC2 * nC2 + nCnC3+.......................nCn * nC =(2nCn)

how u calculate this?

7

Proof

(1+x)n=nC0.x0+nC1.x1+nC2.x2+nC3.x3+......nCn.xn   ............equation 1

(x+1)n=nC0.xn+nC1.xn-1+nC2.xn-2+nC3.xn-3+.......nCn.xn-n ......equation 2

multiply equation 1 and 2 and compare coefficient of xn in LHS and RHS

// Coefficient of xr in (1+x)n is nCr

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

0
take 2n=4

n=2 so for good party male and female are equal

so combination like 1-1 , 2-2

2 party possible

now by option D is ans :P
1

if 2n=4 then  $^4C_2=6$  good parties possible.

0
Sounds like someone has taken an excerpt from ROSEN

Nice 1

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

0

Yeah, this the correct way to solve such problems.Thanks:-)

we have to select n females from 2n people

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

so ans is D
1 flag:
✌ Low quality (ken “Approach is wrong”)

edited
1
Why we have to select $n$ females please elaborate .
2
Mam to choose n female from 2n people number of ways is 1 only (given that remaining n people are male)

$\binom{2n}{n}$ tells number of ways for choosing n people from 2n people, and this approach in no ways solves the given problem.

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