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.
Baba said:
1 decade ago
You could also take a shortcut and say that there are 10*9*8*7 / 4! = 210 ways to arrange 10 people in a group of 4. Then subtract the number of ways to arrange 4 girls in a group of 4 4*3*2*1 / 4! = 1. So 210 Total - 1 All Girl = 209 At least 1 Boy.
Jyoti Nagpal said:
1 decade ago
Another Shortcut Method: Total student=6+4=10 The number of ways to selecting 4 girls=10c4, The number of ways to selecting no boys= 4c4 because total boys are=6 so 10-6=4 to select 4 childern so, 4c4 so selecting atleast one boy 10c4-4c4=210-1=209.
Huma_ said:
5 years ago
Can anyone explain this 2nd step:-.
(6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4).
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
Why the nCr formula is applied to only (6C1 x 4C3) & (6C4) and not to all? Please explain.
(6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4).
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
Why the nCr formula is applied to only (6C1 x 4C3) & (6C4) and not to all? Please explain.
(3)
Richa said:
1 decade ago
It can also be done in another way... we can calculate total no of ways to select the children..ie 6+4=10 10C4 and subtract ways in which no boy is there ie all girls 4C4...
10C4-4C4= 209
Calculation would be easy in this way
10C4-4C4= 209
Calculation would be easy in this way
Maggi said:
1 decade ago
In 1st step :........+ (6c1 x 4c3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4).
In 2nd step :........+ (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
If we use formula nC(n-r) then it is not applied to 6C2 x 4C2. So on.
In 2nd step :........+ (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
If we use formula nC(n-r) then it is not applied to 6C2 x 4C2. So on.
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.
Msrinivasarao said:
1 decade ago
We have 6-boys and 4-girls we have to select only 4 members from the group In that at least one boy should be there.so
we cab write
total number of ways-no boy can be selected
10c4-4c4=210-1
=209
we cab write
total number of ways-no boy can be selected
10c4-4c4=210-1
=209
Nishant Khanorkar said:
1 decade ago
Best way to solve such problem is by going the other way round.
At least 1 boy = total - all girls.
So,
10c4-4c4
= (10*9*8*7)/(1*2*3*4) - 1.
= 210 - 1.
= 209.
And here it is. out required answer.
At least 1 boy = total - all girls.
So,
10c4-4c4
= (10*9*8*7)/(1*2*3*4) - 1.
= 210 - 1.
= 209.
And here it is. out required answer.
Sayuj said:
4 years ago
Why can't we just simply do 6c1*9c3?!
In this, we've included that "minimum" condition of including that 1 boy and the rest 9 kids can be selected at random. why not? Someone, please explain.
In this, we've included that "minimum" condition of including that 1 boy and the rest 9 kids can be selected at random. why not? Someone, please explain.
(1)
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.
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