Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 6)
6.
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
Answer: Option
Explanation:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
 Required numberof ways |
= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) | |||||||||||||||||||
| = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2) | ||||||||||||||||||||
|
||||||||||||||||||||
| = (24 + 90 + 80 + 15) | ||||||||||||||||||||
| = 209. |
Discussion:
92 comments Page 4 of 10.
Mayur said:
1 decade ago
One Question has many times came no one gave answer.
Why formula use for only first term ncr(n-r)?
Why not for other one, how can we know when we use this formula?
Why formula use for only first term ncr(n-r)?
Why not for other one, how can we know when we use this formula?
Atul said:
1 decade ago
I am not satisfied from the above ways because I don want to use simpler method.
Tell me why did we apply formula only on 1st and last terms. Why not on other terms?
Tell me why did we apply formula only on 1st and last terms. Why not on other terms?
Sandy said:
7 years ago
@ALL.
So you can select 1 boy in 6C1 ways. So the rest 3 of them can be any gender right. So why can't it be done in 9C3 ways?
Why is the final answer not 6C1*9C3?
So you can select 1 boy in 6C1 ways. So the rest 3 of them can be any gender right. So why can't it be done in 9C3 ways?
Why is the final answer not 6C1*9C3?
Joe said:
1 decade ago
1 boy can be selected in 6 ways.
The rest 3 can be selected from the remaining 9, in 9C3 ways=84 ways.
Hence the total no: of ways = 6*84 = 504.
Is it right?
The rest 3 can be selected from the remaining 9, in 9C3 ways=84 ways.
Hence the total no: of ways = 6*84 = 504.
Is it right?
Tarun said:
8 years ago
@Harika.
It's because those two events are needed to complete the work. If any one event is needed to complete the work, we use addition. Hope this helped.
It's because those two events are needed to complete the work. If any one event is needed to complete the work, we use addition. Hope this helped.
Shubham said:
10 years ago
Guys just calculate the opposite of that, case where no boy is there.
4c4 = 1.
Then all case which are (4+6) c4 = 210.
Then answer is 210-1 = 209.
4c4 = 1.
Then all case which are (4+6) c4 = 210.
Then answer is 210-1 = 209.
ATM said:
2 years ago
A Group of 4 children, out of 10 children will be;
10c4 = 210.
As 4 girls are there;
So only 1 group is possible without the boy
Hence 210-1 = 209.
10c4 = 210.
As 4 girls are there;
So only 1 group is possible without the boy
Hence 210-1 = 209.
(30)
Badri said:
9 years ago
= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
(6C1 x 4C3) AND (6C4)applied to ncr.
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
(6C1 x 4C3) AND (6C4)applied to ncr.
RAVI said:
1 decade ago
Can also be written as:
No.of ways (at least one boy) = Total no.of ways - No.of ways with no boy = 10c4 - 4c4 = 210 - 1 = 209.
No.of ways (at least one boy) = Total no.of ways - No.of ways with no boy = 10c4 - 4c4 = 210 - 1 = 209.
Shivam said:
6 years ago
No boys selected=4c4 ways i.e 1.
Selection of 4 from all =10c4 ways=10*9*8*7/4*3*2*1=210 ways.
Atleast 1 boy=210-1 = 209 ways.
Selection of 4 from all =10c4 ways=10*9*8*7/4*3*2*1=210 ways.
Atleast 1 boy=210-1 = 209 ways.
(3)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers