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 3 of 10.
Himanshu Kulkarni said:
10 years ago
The question says at least one boy is to be selected.
Selecting 1 boy from 6 boys is 6C1 i.e. 6 ways.
And from remaining 9 children 3 are to be selected randomly (any combination of girls and boys) which is 9C3.
So why not 6C1 x 9C3 matches the answer?
Selecting 1 boy from 6 boys is 6C1 i.e. 6 ways.
And from remaining 9 children 3 are to be selected randomly (any combination of girls and boys) which is 9C3.
So why not 6C1 x 9C3 matches the answer?
Amitabha said:
1 decade ago
We have to select 4 children's.
So possibilities, as you have explained, are 1 boy, 3 girls, 2 boys, 2 girls, 3 boys, 1 girl and 4 boys.
But I think we can consider 4 girls also.
So possibilities, as you have explained, are 1 boy, 3 girls, 2 boys, 2 girls, 3 boys, 1 girl and 4 boys.
But I think we can consider 4 girls also.
Kavya said:
1 decade ago
Why there is no 4C4 means 4 girls?
Rishi Jain said:
1 decade ago
Hi @Aditya,
In 1st step : ......+ (6c1 x 4c3) +......
In 2nd step : ......+ (6c1 x 4c1) +...... How ?
You say that,
Its because nCr = nC (n-r).
So 4C3 = 4C (4-3) = 4C1.
Its one of the formula. But I have confused.
That what about 6c1 we should also do 6c1 = 6c (6-1) = 6c5.
But we didn't know why? Please explain me.
In 1st step : ......+ (6c1 x 4c3) +......
In 2nd step : ......+ (6c1 x 4c1) +...... How ?
You say that,
Its because nCr = nC (n-r).
So 4C3 = 4C (4-3) = 4C1.
Its one of the formula. But I have confused.
That what about 6c1 we should also do 6c1 = 6c (6-1) = 6c5.
But we didn't know why? Please explain me.
Sugumar said:
1 decade ago
Why didn't you take 4 girls out of 4 girls?
Mujhse said:
1 decade ago
What if the question was at least 2 boys? Need help.
Shubham said:
1 decade 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.
Manek said:
1 decade ago
I have understood this sum.
Payal said:
1 decade ago
Second step is not clear.
Aarav said:
8 years ago
We can also apply nCr = n!/((r!)(n-r)!).
It's one of the formulae.
It's one of the formulae.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers