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 2 of 10.
Nithin said:
5 years ago
Why can't we do 6C1*9C3?
First choose a boy from group of 6 and then choose any three from remaining 9 people.
First choose a boy from group of 6 and then choose any three from remaining 9 people.
P.krishna reddy said:
5 years ago
Atleast one = Total - None = 10c4-1 = 209.
(4)
Sathvika said:
6 years ago
4c2 = 4!/2!*(4-2)!.
= 4 * 3/2 * 2,
= 3.
= 4 * 3/2 * 2,
= 3.
Dhanraj said:
6 years ago
Thanks for explaining it in detail @Shivam.
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)
Aniket said:
7 years ago
@All.
6c4 = 6c2.
Because 6c4= 6!/(4!*2!) and 6c2=6!/(2!*4!) both are same that's why 6c4=6c2.
6c4 = 6c2.
Because 6c4= 6!/(4!*2!) and 6c2=6!/(2!*4!) both are same that's why 6c4=6c2.
Imnikesh said:
7 years ago
How 6c4 is reduced to 6c2?
Tab said:
7 years ago
Thanks @Udaya.
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?
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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