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 2 of 14.
Olga said:
1 decade ago
I tried to calculate it differently. First, I calculated number of combinations with less than 3 men in them: 7C2*6C3=21*20=420. Then, total number of combinations possible: 13C5=1287. After that I deducted number of combinations with less than 3 men in them from total number of combinations: 1287-420= 862. However the answer is incorrect. Please, where is my mistake?
Pratheeksha said:
8 years ago
Yes, 756 is a right ans.
Bcause out of 13(7men+6women) we have to select 5.Therefore
7men(5)6women,
7C3 X 6C2=35(15)=525,
7C4 X 6C1=35(6)=210,
7C5 X 6C0=21(1)=21,
=756.
Therefore, 756 is the answer.
Bcause out of 13(7men+6women) we have to select 5.Therefore
7men(5)6women,
7C3 X 6C2=35(15)=525,
7C4 X 6C1=35(6)=210,
7C5 X 6C0=21(1)=21,
=756.
Therefore, 756 is the 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.
Mohan said:
1 decade ago
@Nikhil,
The combinations in 7C3 and 10C2 are not mutually exclusive. Hence we cannot just multiply both and result in a solution.
Its like saying, the combinations that 7C3 gives is same as what
7C1*6C1*5C1 would give.(selecting 1 out of 7 as first Man, 1 out of remaining 6 men for second...etc)
That is 35 is not equal to 210.
The combinations in 7C3 and 10C2 are not mutually exclusive. Hence we cannot just multiply both and result in a solution.
Its like saying, the combinations that 7C3 gives is same as what
7C1*6C1*5C1 would give.(selecting 1 out of 7 as first Man, 1 out of remaining 6 men for second...etc)
That is 35 is not equal to 210.
(1)
VAMSI REDDY 1100 said:
1 decade ago
Hey guys!
I am Vamsi I got a small doubt.
Please clear my doubt.
For example if we want to choose 3 items from a list of 10 items, can the answer be c(10,1)*c(9,1)*c(8,1) instead of c(10,3).
My interpretation was I broke down the process of choosing into choosing 1 thing for 3 times...and then apply product rule.
Please help me.
I am Vamsi I got a small doubt.
Please clear my doubt.
For example if we want to choose 3 items from a list of 10 items, can the answer be c(10,1)*c(9,1)*c(8,1) instead of c(10,3).
My interpretation was I broke down the process of choosing into choosing 1 thing for 3 times...and then apply product rule.
Please help me.
Ankz said:
1 decade ago
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 10C2. Please Clarify.
Anyone please elaborate whats wrong in this approach ?
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 10C2. Please Clarify.
Anyone please elaborate whats wrong in this approach ?
Shams said:
1 decade ago
@Riya.
This is a Problem of Combination, There is only one formula is applied that is,
nCr = n!/(n-r)!r!.
(3 men and 2 women).
For man, n=7 & r=3.
For women n=6 & r=2.
(7C3 x 6C2) = 7! / (7-3)! 3! = 7*6*5*4*3! /4*3*2*1*3! = 840/24.
= 35.
So using this formula proceed further steps to get the result.
This is a Problem of Combination, There is only one formula is applied that is,
nCr = n!/(n-r)!r!.
(3 men and 2 women).
For man, n=7 & r=3.
For women n=6 & r=2.
(7C3 x 6C2) = 7! / (7-3)! 3! = 7*6*5*4*3! /4*3*2*1*3! = 840/24.
= 35.
So using this formula proceed further steps to get the result.
Pandu said:
1 decade ago
@Priyanka das.
7C2 has taken since in the last case we need to take 5 men so that you may be understood, its actually 7c5 when in nCr if n value is more than half of n value we need to use formula as nC(n-r)so implies n=7, r=5 since r is more than half of n we used as 7C(7-5) it becomes 7C2. OK I HOPE you got it.
7C2 has taken since in the last case we need to take 5 men so that you may be understood, its actually 7c5 when in nCr if n value is more than half of n value we need to use formula as nC(n-r)so implies n=7, r=5 since r is more than half of n we used as 7C(7-5) it becomes 7C2. OK I HOPE you got it.
Mamta Patel said:
8 years ago
The total number of combinations possible: 13C5=1287.
Then Calculate the number of combinations with less than 3 men in them: 7C2*6C3=21*20=420 7C1*6C4=7*15=105 7C0*6C5=1*6=6.
After that deduct the number of combinations with less than 3 men from total number of combinations: 1287-(420+105+6)= 756.
Then Calculate the number of combinations with less than 3 men in them: 7C2*6C3=21*20=420 7C1*6C4=7*15=105 7C0*6C5=1*6=6.
After that deduct the number of combinations with less than 3 men from total number of combinations: 1287-(420+105+6)= 756.
Shirin Sultana said:
7 years ago
But many ones falling problems not finding the details solution of the combination. It might be done more details as like:
7C4 = 7!/4!*(7-4)!
= 7*6*5*4!/4!*3*2*1.
= 7*6*5/3*2*1,
= 35.
And that would be understandable to all.
By the by, May I know the basic differences between combination and permutation.
7C4 = 7!/4!*(7-4)!
= 7*6*5*4!/4!*3*2*1.
= 7*6*5/3*2*1,
= 35.
And that would be understandable to all.
By the by, May I know the basic differences between combination and permutation.
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