Aptitude - Permutation and Combination - Discussion

Men=7, Women = 6.

Method -1 (correct).

First of all, we need to select 3 men from 7 so = 7C3 = 35.
In problem, they are asking for a minimum of 3 men so already you took 3 men.

So now we need women.
So we need 2 women from 6 women.
So 6c2.
So the value of 6c2 is 15.
Now 35*15 = 525.

In same way, the second committee we took 4 men from the 7 men group.

i.e 7c4 = 35.

We need one more woman for the 2nd committee because already we took 4 men.

So, 6c1 = 6.
So, 35*6 = 210.

Now we took all men for 3 rd committee.
So 7c5 = 21.

Now the total is 525+210+21 = 756.


METHOD II (Seems to be correct).

Why would the logic 7C3 * 10C2 not work? First, we've chosen 3 Men - 7C3. Now we have 10 people left (which included both men and women). Then we need to select any 2 people either men or women so 10C2.

Mistake: let 7 men be a, b, c, d, e, f, g.

7c3==> (a, b, c) ; (b, c, d) ; (c, d, e) ;. 35 sets.

Let us look into one condition.

4 men, 1women (let us assume that women is fixed).

(a b c) ;d;fixed women.
(b c d) a; fixed women.
(a b d);c;fixed women.

And so in a similar way, there will be the same items counted multiple times, in method 1 this does not occur.

@Vamsi.

As @Mohan said to do multiplication A * C for instance event A shld not in any way affect the occurrence of C.

Now from a list of 10 if 1 item is selected then that item can never be selected from the list of next 9 items AND HERE IS THE CATCH so removing any specific item in c(10,1) will make sure that specific item will not be selected in c(9,1).

So c(10,1) and c(9,1) is not mutually exclusive since they
both remove elements from same list and WE CAN NOT MULTIPLY
NON MUTUALLY EXCLUSIVE events.

Let assume, there are three nodes STATION 1 , STATION 2, STATION 3.

If we have 3 path from STATION 1 to STATION 2 and 3 path from STATION 2 to STATION3 then we have 9 path from station 1 to station3, but think if taking any from station 1 to station 2 causes blockage of any one path from station 2 to station 3.

Then total number of path from station 1 to station 3 is 5.

Hope you got it :).

Basic Formula is nPr=n!/(n-r)!
n=number
r=spaces

And in case of repetition(if number/character is repeating) we have to apply this formula :nPr/(No .of repeated number/character)! ==>n!/(n-r)!/(No .of repeated number/character)!

Based On Question :

We got three possibilities:

First possibility:- (7 men 6 women) out of which we can select (3 men 2 women)
=>>(7P3/3! * 6P2/2!)
==>(35*15)
{men repeated 3 times and women 2 times that's why i have take 3! and 2!}
==>525----> (1)

Second possibility:-(7 men 6 women) out of which we can select (4 men 1 women)
==>(7P4/4! * 6P1/1!)==>35*6 = 210-----> (2)

Third possibility;-(7 men:6) out of which we can select (5 men only )
==>7P5/5! = 21----> (3)

Add all equatations (1),(2) and (3)
==>525 + 210 + 21 = 756.

In that question that is cleared that 5 members need to be select out of 7 men and 6 women.

And in question there is a condition that at least 3 men that mean we can select in 3 ways that we can select 3 men or 4 men or 5 men.

1) 3 men and we want another 5-3 = 2 women. We have to select 3 men from 7 men that is 7C3 and 2 womens from 6 women that is 6C2 results is 7C3*6C2.

2) 4men and we want another 5-4 = 1 women. We have to select 4 men from 7 men that is 7C4 and 1 women from 6 women that is 6C1 results is 7C4*6C1.

3) 5 men and we want another 5-5 = 0 women. We have to select 5 men from 7 men that is 7C5 and 0 women from 6 women that is 6C2 results is 7C5*6C0.

Answer is we need to sum up all these combinations that is (7C3*6C2+7C4*6C1+7C5*6C0) that is 756.

Yes it's write answer.

Using combination formula.
nCr = n! / r! ( n - r ) !
Question says at least 3 men's , means we have to select minimum 3 men's , it means we can select either 3 men's or 4 men's or 5 men's.
1)in first case select 3 men's and 2 women's. To form 5 people commette ..
7! / 3! (7-3)! * 6! / 2!(6-2)! - Eq-1
2 ) in second case select 4 men's and 1 women can be selected in.
7! / 4!(7-4)! * 6! / 1!(6-1)! - Eq -2
3) in third cas we can select 5 men's only
Can be selected in 7! / 5!(7-5)! - Eq -3

The final answer is:

Eq-1 + Eq-2 + Eq-3=
7! / 3! (7-3)! * 6! / 2!(6-2)! + 7! / 4!(7-4)! * 6! / 1!(6-1)! + 7! / 5!(7-5)!
= 525 + 210 + 21
=756 ways.

The question is not specific regarding the selection of men and women after the selection of at least three men.

If we take the method [7C3 * 6C2] + [7C4 * 6C1] + [7C5 * 6C0] as the solution, then we must understand that all men are taken as one entity first then women are considered as another entity.

Hence, 3 men are one entity and 2 women are another entity. Similarly, 4 men are one and likewise, 5 men are one entity.

That means if you are done with the selection of 3 men at least, then DONT SELECT MEN AGAIN from the remaining. And if you choose to select men again, THEN INCLUDE THEM WHILE SELECTING ALTEAST THREE MEN.

Hi Naveen,

you are wrong

first step is correct that is
First of all we need to select 3 man from 7 so = 7C3 = 35
and second step you are wrong

in problem they are asking minimum of 3 men so already you took 3 men
so now we need woman

so we need 2 women from 6 women

so 6c2
so the value of 6c2 is 15

now 35*15 = 525

like same way second committee we took 4 men from 7 men group
i.e. 7c4 = 35

we need one more woman for 2nd committe because already we took 4 men

so 6c1 = 6

so 35*6 = 210

now we took all men for 3 rd commitee so 7c5 = 21

now the total is 525+210+21 = 756

I hope you have understood the problem.

I am repeating @Nikhil and @Ankzs and my query here:

I think @Nikhil's query still not answered

Why would the logic 7C3 * 10C2 not work? First we've choosen 3 Men - 7C3. Now we have 10 people left (which included both men and women). Then we need to select any 2 people either men or women so (6+4=10)C2. Please Clarify.

Anyone please elaborate what's wrong in this approach ?

Why do we need to separately consider cases in which there are 3 men, 4 men and 5 men?

10C2, [ having 4 men and 6 women ] includes the possibilities of selection of 4 guys and 5 guys.

Hi,

Could you please explain in what condition or scenario the formula nCr=nCn-r will implicated.

Its used in (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
= ......... + (7C3 x 6C1) + (7C2)ncr= ncn-r

With this formula I came to know that 7c4 and 7c5 will become 7C3 and 7c2 respectively. But I could not understand why this formula is used here, Why cant we use the formula ncr=n!/r!(n-r)!.

Pleas anybody kindly help me in clearing the above doubt.

Thanks in advance.

You can't use 7C3*10C2.
Because suppose the 5 men are ABCDE and 2 women are XY.

If you say 7C3*10C2 let the men be ABC and the other two be DX
then the committee will be ABCDX.

If you consider another way for 7C3*10C2 let da men be ABD and the other two be CX.

Now the committee is ABDCX.

But ABCDX = ABDCX.

But by using 7C3*10C5 it is taken as two separate ways.. but it is one. So that method is wrong :).