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:
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?
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.
Kavya said:
9 years ago
Why there is no 4C4 means 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.
Sugumar said:
10 years ago
Why didn't you take 4 girls out of 4 girls?
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.
Manek said:
10 years ago
I have understood this sum.
Rahul said:
10 years ago
Why we are not using nc(n-r) formula in all in 2nd step?
Heenu said:
7 years ago
Easiest way:
Since 4 students are to be selected and there are 4 girls, hence we don't want a combination consisting of all girls: 4c4 =1.
Total no of combinations available: (Total students= 6 boys + 4 girls =10 and 4 need to be selected) 10c4= 210.
Hence favorable combinations = 210 -1 = 209.
Since 4 students are to be selected and there are 4 girls, hence we don't want a combination consisting of all girls: 4c4 =1.
Total no of combinations available: (Total students= 6 boys + 4 girls =10 and 4 need to be selected) 10c4= 210.
Hence favorable combinations = 210 -1 = 209.
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