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 |
= (7C3 x 6C2) + (7C4 x 6C1) + (7C5) | |||||||||||
|
||||||||||||
|
||||||||||||
| = (525 + 210 + 21) | ||||||||||||
| = 756. |
Discussion:
137 comments Page 4 of 14.
Rakesh said:
1 decade ago
= (7C3 x 6C2) + (7C4 x 6C1) + (7C5).
= 7 x 6 x 5x6 x 5+ (7C3 x 6C1) + (7C2).
3 x 2 x 1 2 x 1.
From 1st to 2nd line I have not understood please ex-plane.
How (7C4 x 6C1) + (7C5) become (7C3 x 6C1) + (7C2)?
= 7 x 6 x 5x6 x 5+ (7C3 x 6C1) + (7C2).
3 x 2 x 1 2 x 1.
From 1st to 2nd line I have not understood please ex-plane.
How (7C4 x 6C1) + (7C5) become (7C3 x 6C1) + (7C2)?
Zara said:
10 years ago
I used combination formula, nCr = n!/r! (n-r)!
Required number of ways = (7C3*6C2)+(7C4*6C1)+7C5.
= (7*6*5/3*2*1+6*5*1/2*1)+(7*6*5*4 /4*3*2*1+6/1)+(7*6*5*4*3*2*1/5*4*3*2*1).
= 35*15+35*6+24 = 756 answer.
Required number of ways = (7C3*6C2)+(7C4*6C1)+7C5.
= (7*6*5/3*2*1+6*5*1/2*1)+(7*6*5*4 /4*3*2*1+6/1)+(7*6*5*4*3*2*1/5*4*3*2*1).
= 35*15+35*6+24 = 756 answer.
VATSAL KHANDELWAL said:
1 decade ago
Dear @Akash,
You mustn't try these if you don't even know the meaning of "c".
Else what I can say is that.
"c" simply means "combination".
But you can't understand if you haven't read the chapter.
You mustn't try these if you don't even know the meaning of "c".
Else what I can say is that.
"c" simply means "combination".
But you can't understand if you haven't read the chapter.
Naveen said:
2 decades ago
First of all we need to select 3 man from 7 so = 7C3 = 35
Then we can choose 2 person from (7-3)+ 6 person ie. 10C2 = 45
So answer should be 53*45 = none of these..
Please tell me where I'm wrong.
Then we can choose 2 person from (7-3)+ 6 person ie. 10C2 = 45
So answer should be 53*45 = none of these..
Please tell me where I'm wrong.
Madhusudan said:
1 decade ago
In Above solution I am unable to understand why have we included 5/1(in both the committee and 6/2 in 2nd committee. Can any help me to understand how the combination considered in the example.
Simran said:
9 years ago
I understood all the combination how we choose, Thanks for this @ Mohan & @Pavz. :)
But I have little doubt in solving cases like;
1) why we "multiple" in case?
2) why we "add" all cases?
But I have little doubt in solving cases like;
1) why we "multiple" in case?
2) why we "add" all cases?
Dadasaheb Maske said:
7 years ago
Among 5 children there are 2siblings.in how many ways children be seated in a row so that the siblings do not sit together.
A) 86
B) 72
C) 46
D) 38
How to solve it? Please anyone explain me.
A) 86
B) 72
C) 46
D) 38
How to solve it? Please anyone explain me.
(1)
Dimi said:
7 years ago
I have a question, in how many ways can a committee of nine people can be formed from ten men and their wives. If no man is to serve on the committee with his wife?
Please solve this.
Please solve this.
Rutuja said:
9 years ago
Someone, please give the solution.
In a group of 6 people, there are 3 Indians and 3 Chinese. How many subsets can be created such that there are at least 1 Indian in each subset?
In a group of 6 people, there are 3 Indians and 3 Chinese. How many subsets can be created such that there are at least 1 Indian in each subset?
Nishi Kant said:
1 decade ago
Here it is clearly given that we have to select so here we have to apply the combination rule and at-least three is given so we have to start from three and proceed further.
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