Aptitude  Permutation and Combination  Discussion
Discussion Forum : Permutation and Combination  General Questions (Q.No. 1)
1.
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
Answer: Option
Explanation:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways  = (^{7}C_{3} x ^{6}C_{2}) + (^{7}C_{4} x ^{6}C_{1}) + (^{7}C_{5})  




= (525 + 210 + 21)  
= 756. 
Discussion:
131 comments Page 1 of 14.
Sumanth reddy said:
6 months ago
How it's changed from 7c4 to 7c3? Please explain me.
(8)
Suga said:
2 years ago
@All.
Here, some of the persons are confused in the second step because there was an incorrection where (7C4 x 6C1) + (7C5) were converted into + (7C3 x 6C1) + (7C2).
Here, some of the persons are confused in the second step because there was an incorrection where (7C4 x 6C1) + (7C5) were converted into + (7C3 x 6C1) + (7C2).
(7)
Abi said:
2 years ago
Anyone, Please explain this 7C3 * 6C1+7C2.
(7)
Y Krishna Pavan Sai said:
2 years ago
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.
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.
(10)
Anu said:
2 years ago
I haven't understood that point. Anyone, please help me to get it.
(3)
Johar said:
2 years ago
Anyone, Please solve this problem clearly to get it.
(1)
Jay said:
2 years ago
@Nivetha.
We only have to find total combinations, not the probability. If we want to find probability then we have to divide it by 13C5.
We only have to find total combinations, not the probability. If we want to find probability then we have to divide it by 13C5.
(2)
Shaifali said:
3 years ago
Well explained @Nagu.
But I don't understand why do we need to multiply? why can't we add?
But I don't understand why do we need to multiply? why can't we add?
Sake said:
3 years ago
Why can't we do it like this,
Total ways = at least 3 men + not men.
i.e not men = 6c5.
Total ways= 13c5.
therefore,
13c5  6c5.
Total ways = at least 3 men + not men.
i.e not men = 6c5.
Total ways= 13c5.
therefore,
13c5  6c5.
(2)
Akshay said:
4 years ago
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! (73)! * 6! / 2!(62)!  Eq1
2 ) in second case select 4 men's and 1 women can be selected in.
7! / 4!(74)! * 6! / 1!(61)!  Eq 2
3) in third cas we can select 5 men's only
Can be selected in 7! / 5!(75)!  Eq 3
The final answer is:
Eq1 + Eq2 + Eq3=
7! / 3! (73)! * 6! / 2!(62)! + 7! / 4!(74)! * 6! / 1!(61)! + 7! / 5!(75)!
= 525 + 210 + 21
=756 ways.
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! (73)! * 6! / 2!(62)!  Eq1
2 ) in second case select 4 men's and 1 women can be selected in.
7! / 4!(74)! * 6! / 1!(61)!  Eq 2
3) in third cas we can select 5 men's only
Can be selected in 7! / 5!(75)!  Eq 3
The final answer is:
Eq1 + Eq2 + Eq3=
7! / 3! (73)! * 6! / 2!(62)! + 7! / 4!(74)! * 6! / 1!(61)! + 7! / 5!(75)!
= 525 + 210 + 21
=756 ways.
(3)
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