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 5 of 10.
Tarun said:
9 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.
Harika said:
9 years ago
I'm bit confused that why here use multiplication? Please tell me.
Pranali said:
9 years ago
In first step (6c2 * 4c2).
Send step ( (6*5/2*1) * (4*3/2*1) ).
How?
I can't understand it
Send step ( (6*5/2*1) * (4*3/2*1) ).
How?
I can't understand it
Shivaraj said:
9 years ago
@Neema.
It's mentioned, read question clearly.
It's mentioned, read question clearly.
DANIEL said:
9 years ago
Why not 4 girls?
Neema said:
9 years ago
Not mention that how many person should we take at a time, the how it will be solved?
Pallavi said:
1 decade ago
Another easy method is
1-(4C4/10C4)
=1-(1/210)
=209/210
Therefore 209 times out of possible 210.
1-(4C4/10C4)
=1-(1/210)
=209/210
Therefore 209 times out of possible 210.
Anusha said:
1 decade ago
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
Why to change only the first term and last term with formula. Why don't the formula apply to the other terms. Please Explain?
Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
Why to change only the first term and last term with formula. Why don't the formula apply to the other terms. Please Explain?
Kashika shekhar said:
1 decade ago
I am not able to apply different types of logic in different questions. Please tell me some logic for appying the formulas.
Shanmuga kumar said:
1 decade ago
How do you get 4 members in group? please explain.
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