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 10 of 14.
Mayur said:
1 decade ago
Why only that formula used in 7c4 n 7c5. ncr=nc (n-r)?
Why didn't we use in 6c1? Can anybody clear me please?
Why didn't we use in 6c1? Can anybody clear me please?
Ananya said:
1 decade ago
Is there any shortcut for this?
Naji said:
1 decade ago
How can we notice that we just have to select without arrangement ?
Sam said:
1 decade ago
Can you please explain 7c5 is converted into nc(n-r)?
But why didn't convert 6c2 into 6c(6-2)?
But why didn't convert 6c2 into 6c(6-2)?
Naveen Kumar said:
1 decade ago
Why we are calculating 7C5 as 7C2?
Maryann said:
1 decade ago
Can someone explain better for me because I'm getting confused?
Ajay said:
1 decade ago
7C5 written as 7C2 because 7C5 = 7C(7-5) both having same values from formula nCr = nC(nr).
Ajaykumar said:
1 decade ago
7C3 = 35, 7!/(7-3)!3! = 7!/4!3! = 7*6*5/3*2*1 = 35.
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.
Trinity said:
10 years ago
Why 7C3*10C2 is wrong?
Lets take a case:
We have 7 men namely A B C D E F G and 6 women H I J K L M.
Case 1) 3 men and 2 from the set of 10 = A B C + D H.
Case 2) 3 men and 2 from the set of 10 = A B D + C H.
Both the sets are identical. This shows that our sets are not mutually exclusive. Hence we must choose men and women separately.
Lets take a case:
We have 7 men namely A B C D E F G and 6 women H I J K L M.
Case 1) 3 men and 2 from the set of 10 = A B C + D H.
Case 2) 3 men and 2 from the set of 10 = A B D + C H.
Both the sets are identical. This shows that our sets are not mutually exclusive. Hence we must choose men and women separately.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers