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 6 of 10.
Mujhse said:
10 years ago
What if the question was at least 2 boys? Need help.
Sugumar said:
10 years ago
Why didn't you take 4 girls out of 4 girls?
Rishi Jain said:
10 years 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.
Kavya said:
10 years ago
Why there is no 4C4 means 4 girls?
Amitabha said:
9 years 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.
Himanshu Kulkarni said:
9 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?
Chiru said:
9 years ago
Its easy.
10c4 - 4c4 =209.
10c4 - 4c4 =209.
Sharad said:
9 years ago
@Udaya.
May I know how could you find out the answer from below calculation?
So the number of ways the selection can be made such that at least one boy should be there is =10C4 - 4C4 = 209.
Please explain this.
May I know how could you find out the answer from below calculation?
So the number of ways the selection can be made such that at least one boy should be there is =10C4 - 4C4 = 209.
Please explain this.
Sonu said:
9 years ago
Should not we have to arrange all the children? In this we have just make combinations.
Kayalvizhi said:
9 years ago
Can anyone explain how to derive this calculation? Please.
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